zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

1+sin^2(x)+cos^4(x)=0

解答

1 + sin^{2} (x) + cos^{4} (x) = 0

解答

x \in R 无解

求解步骤

1 + sin^{2} (x) + cos^{4} (x) = 0

使用三角恒等式改写

1 + cos^{4} (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + cos^{4} (x) + 1 - cos^{2} (x)

化简

1 + cos^{4} (x) + 1 - cos^{2} (x) : cos^{4} (x) - cos^{2} (x) + 2

1 + cos^{4} (x) + 1 - cos^{2} (x)

对同类项分组

= cos^{4} (x) - cos^{2} (x) + 1 + 1

数字相加:

1 + 1 = 2

= cos^{4} (x) - cos^{2} (x) + 2

= cos^{4} (x) - cos^{2} (x) + 2

2 - cos^{2} (x) + cos^{4} (x) = 0

用替代法求解

2 - cos^{2} (x) + cos^{4} (x) = 0

令

: cos (x) = u

2 - u^{2} + u^{4} = 0

2 - u^{2} + u^{4} = 0 : u = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i, u = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

2 - u^{2} + u^{4} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - u^{2} + 2 = 0

用

a = u^{2}

和

a^{2} = u^{4}

改写方程式

a^{2} - a + 2 = 0

解

a^{2} - a + 2 = 0 : a = \frac{1}{2} + i \frac{7}{2}, a = \frac{1}{2} - i \frac{7}{2}

a^{2} - a + 2 = 0

使用求根公式求解

a^{2} - a + 2 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = 2

a_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

a_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

化简

(- 1)^{2} - 4 \cdot 1 \cdot 2 : 7 i

(- 1)^{2} - 4 \cdot 1 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 - 8

数字相减:

1 - 8 = - 7

= - 7

使用根式运算法则:

- a = - 1 a

- 7 = - 1 7

= - 1 7

使用虚数运算法则:

- 1 = i

= 7 i

a_{1, 2} = \frac{- ( - 1 ) \pm 7 i}{2 \cdot 1}

将解分隔开

a_{1} = \frac{- ( - 1 ) + 7 i}{2 \cdot 1}, a_{2} = \frac{- ( - 1 ) - 7 i}{2 \cdot 1}

a = \frac{- ( - 1 ) + 7 i}{2 \cdot 1} : \frac{1}{2} + i \frac{7}{2}

\frac{- ( - 1 ) + 7 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 7 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 7 i}{2}

将

\frac{1 + 7 i}{2}

改写成标准复数形式:

\frac{1}{2} + \frac{7}{2} i

\frac{1 + 7 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + 7 i}{2} = \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7}{2} i

a = \frac{- ( - 1 ) - 7 i}{2 \cdot 1} : \frac{1}{2} - i \frac{7}{2}

\frac{- ( - 1 ) - 7 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 7 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 7 i}{2}

将

\frac{1 - 7 i}{2}

改写成标准复数形式:

\frac{1}{2} - \frac{7}{2} i

\frac{1 - 7 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 - 7 i}{2} = \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7}{2} i

二次方程组的解是:

a = \frac{1}{2} + i \frac{7}{2}, a = \frac{1}{2} - i \frac{7}{2}

a = \frac{1}{2} + i \frac{7}{2}, a = \frac{1}{2} - i \frac{7}{2}

代回

a = u^{2}

，求解

u

解

u^{2} = \frac{1}{2} + i \frac{7}{2} : u = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i

u^{2} = \frac{1}{2} + i \frac{7}{2}

替代

u = a + bi

(a + bi)^{2} = \frac{1}{2} + i \frac{7}{2}

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = \frac{1}{2} + i \frac{7}{2}

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{7}{2}]

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{7}{2}] : a = \frac{7}{2 2 2 - 1}, a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = \frac{2 2 - 1}{2} b = - \frac{2 2 - 1}{2}

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{7}{2}]

对于

2 ab = \frac{7}{2}

将

a

移到一边:

a = \frac{7}{4 b}

2 ab = \frac{7}{2}

两边除以

2 b

2 ab = \frac{7}{2}

两边除以

2 b

\frac{2 ab}{2 b} = \frac{\frac{7}{2}}{2 b}

化简

\frac{2 ab}{2 b} = \frac{\frac{7}{2}}{2 b}

化简

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

数字相除:

\frac{2}{2} = 1

= \frac{ab}{b}

约分:

b

= a

化简

\frac{\frac{7}{2}}{2 b} : \frac{7}{4 b}

\frac{\frac{7}{2}}{2 b}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7}{2 \cdot 2 b}

数字相乘:

2 \cdot 2 = 4

= \frac{7}{4 b}

a = \frac{7}{4 b}

a = \frac{7}{4 b}

a = \frac{7}{4 b}

将解

a = \frac{7}{4 b}

代入

a^{2} - b^{2} = \frac{1}{2}

对于

a^{2} - b^{2} = \frac{1}{2}

，用

\frac{7}{4 b}

替代

a : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

对于

a^{2} - b^{2} = \frac{1}{2}

，用

\frac{7}{4 b}

替代

a

(\frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

解

(\frac{7}{4 b})^{2} - b^{2} = \frac{1}{2} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

(\frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

乘以最小公倍数

(\frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

化简

(\frac{7}{4 b})^{2} : \frac{7}{16 b ^{2}}

(\frac{7}{4 b})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 7 ) ^{2}}{( 4 b ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(4 b)^{2} = 4^{2} b^{2}

= \frac{( 7 ) ^{2}}{4 ^{2} b ^{2}}

(7)^{2} : 7

使用根式运算法则:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 7^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 7

= \frac{7}{4 ^{2} b ^{2}}

4^{2} = 16

= \frac{7}{16 b ^{2}}

\frac{7}{16 b ^{2}} - b^{2} = \frac{1}{2}

找到

16 b^{2}, 2

的最小公倍数:

16 b^{2}

16 b^{2}, 2

最小公倍数 (LCM)

16, 2

的最小公倍数:

16

16, 2

最小公倍数 (LCM)

16

质因数分解:

2 \cdot 2 \cdot 2 \cdot 2

16

16

除以

216 = 8 \cdot 2

= 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

16

或

2

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

计算出由出现在

16 b^{2}

或

2

中的因子组成的表达式

= 16 b^{2}

乘以最小公倍数=

16 b^{2}

\frac{7}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{7}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{7}{16 b ^{2}} \cdot 16 b^{2} : 7

\frac{7}{16 b ^{2}} \cdot 16 b^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{7 \cdot 16 b ^{2}}{16 b ^{2}}

约分:

16

= \frac{7 b ^{2}}{b ^{2}}

约分:

b^{2}

= 7

化简

- b^{2} \cdot 16 b^{2} : - 16 b^{4}

- b^{2} \cdot 16 b^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 16 b^{4}

化简

\frac{1}{2} \cdot 16 b^{2} : 8 b^{2}

\frac{1}{2} \cdot 16 b^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 16}{2} b^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

数字相乘:

1 \cdot 16 = 16

= \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= 8

= 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

解

7 - 16 b^{4} = 8 b^{2} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

7 - 16 b^{4} = 8 b^{2}

将

8 b^{2}

para o lado esquerdo

7 - 16 b^{4} = 8 b^{2}

两边减去

8 b^{2}

7 - 16 b^{4} - 8 b^{2} = 8 b^{2} - 8 b^{2}

化简

7 - 16 b^{4} - 8 b^{2} = 0

7 - 16 b^{4} - 8 b^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 b^{4} - 8 b^{2} + 7 = 0

用

u = b^{2}

和

u^{2} = b^{4}

改写方程式

- 16 u^{2} - 8 u + 7 = 0

解

- 16 u^{2} - 8 u + 7 = 0 : u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

- 16 u^{2} - 8 u + 7 = 0

使用求根公式求解

- 16 u^{2} - 8 u + 7 = 0

二次方程求根公式:

若

a = - 16, b = - 8, c = 7

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 7}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 7}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 7 = 162

(- 8)^{2} - 4 (- 16) \cdot 7

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 7

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 7

数字相乘:

4 \cdot 16 \cdot 7 = 448

= 8^{2} + 448

8^{2} = 64

= 64 + 448

数字相加:

64 + 448 = 512

= 512

512

质因数分解:

2^{9}

512

512

除以

2512 = 256 \cdot 2

= 2 \cdot 256

256

除以

2256 = 128 \cdot 2

= 2 \cdot 2 \cdot 128

128

除以

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 64

64

除以

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{9}

= 2^{9}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{8} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{8}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 2

整理后得

= 162

u_{1, 2} = \frac{- ( - 8 ) \pm 16 2}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 16 2}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16 2}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16 2}{2 ( - 16 )} : - \frac{1 + 2 2}{4}

\frac{- ( - 8 ) + 16 2}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 + 16 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 + 16 2}{- 32}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8 + 16 2}{32}

消掉

\frac{8 + 16 2}{32} : \frac{1 + 2 2}{4}

\frac{8 + 16 2}{32}

分解

8 + 162 : 8 (1 + 22)

8 + 162

改写为

= 8 \cdot 1 + 8 \cdot 22

因式分解出通项

8

= 8 (1 + 22)

= \frac{8 ( 1 + 2 2 )}{32}

约分:

8

= \frac{1 + 2 2}{4}

= - \frac{1 + 2 2}{4}

u = \frac{- ( - 8 ) - 16 2}{2 ( - 16 )} : \frac{2 2 - 1}{4}

\frac{- ( - 8 ) - 16 2}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 - 16 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 - 16 2}{- 32}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

8 - 162 = - (162 - 8)

= \frac{16 2 - 8}{32}

分解

162 - 8 : 8 (22 - 1)

162 - 8

改写为

= 8 \cdot 22 - 8 \cdot 1

因式分解出通项

8

= 8 (22 - 1)

= \frac{8 ( 2 2 - 1 )}{32}

约分:

8

= \frac{2 2 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

代回

u = b^{2}

，求解

b

解

b^{2} = - \frac{1 + 2 2}{4} : b \in R

无解

b^{2} = - \frac{1 + 2 2}{4}

x^{2}

在

x

内不能为负

\in R

b \in R 无解

解

b^{2} = \frac{2 2 - 1}{4} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

b^{2} = \frac{2 2 - 1}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

b = \frac{2 2 - 1}{4}, b = - \frac{2 2 - 1}{4}

\frac{2 2 - 1}{4} = \frac{2 2 - 1}{2}

\frac{2 2 - 1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{2 2 - 1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{2 2 - 1}{2}

- \frac{2 2 - 1}{4} = - \frac{2 2 - 1}{2}

- \frac{2 2 - 1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{2 2 - 1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{2 2 - 1}{2}

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

解为

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(\frac{7}{4 b})^{2} - b^{2}

的分母，令其等于零

解

4 b = 0 : b = 0

4 b = 0

两边除以

4

4 b = 0

两边除以

4

\frac{4 b}{4} = \frac{0}{4}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

将解

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

代入

2 ab = \frac{7}{2}

对于

2 ab = \frac{7}{2}

，用

\frac{2 2 - 1}{2}

替代

b : a = \frac{7}{2 2 2 - 1}

对于

2 ab = \frac{7}{2}

，用

\frac{2 2 - 1}{2}

替代

b

2 a \frac{2 2 - 1}{2} = \frac{7}{2}

解

2 a \frac{2 2 - 1}{2} = \frac{7}{2} : a = \frac{7}{2 2 2 - 1}

2 a \frac{2 2 - 1}{2} = \frac{7}{2}

在两边乘以

2

2 a \frac{2 2 - 1}{2} = \frac{7}{2}

在两边乘以

2

2 \cdot 2 a \frac{2 2 - 1}{2} = \frac{2 7}{2}

化简

2 a 22 - 1 = 7

2 a 22 - 1 = 7

两边除以

2 22 - 1

2 a 22 - 1 = 7

两边除以

2 22 - 1

\frac{2 a 2 2 - 1}{2 2 2 - 1} = \frac{7}{2 2 2 - 1}

化简

a = \frac{7}{2 2 2 - 1}

a = \frac{7}{2 2 2 - 1}

对于

2 ab = \frac{7}{2}

，用

- \frac{2 2 - 1}{2}

替代

b : a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

对于

2 ab = \frac{7}{2}

，用

- \frac{2 2 - 1}{2}

替代

b

2 a (- \frac{2 2 - 1}{2}) = \frac{7}{2}

解

2 a (- \frac{2 2 - 1}{2}) = \frac{7}{2} : a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

2 a (- \frac{2 2 - 1}{2}) = \frac{7}{2}

两边除以

2 (- \frac{2 2 - 1}{2})

2 a (- \frac{2 2 - 1}{2}) = \frac{7}{2}

两边除以

2 (- \frac{2 2 - 1}{2})

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} = \frac{\frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} = \frac{\frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} : a

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} : \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 a (- \frac{2 2 - 1}{2}) = - 2 a \frac{2 2 - 1}{2}

= \frac{- 2 a \frac{2 2 - 1}{2}}{2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{2 2 - 1}{2}) = - 2 \cdot \frac{2 2 - 1}{2}

= \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

= \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

约分:

- 2

= \frac{a \frac{2 2 - 1}{2}}{\frac{2 2 - 1}{2}}

约分:

\frac{2 2 - 1}{2}

= a

化简

\frac{\frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )} : - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

\frac{\frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{7}{2 \cdot 2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 \cdot 2 (- \frac{2 2 - 1}{2}) = - 2 \cdot 2 \cdot \frac{2 2 - 1}{2}

= \frac{7}{- 2 \cdot 2 \cdot \frac{2 2 - 1}{2}}

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= \frac{7}{- 2 ^{2} \cdot \frac{2 2 - 1}{2}}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}

将解代入原方程进行验证

将它们代入

a^{2} - b^{2} = \frac{1}{2}

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = - \frac{2 2 - 1}{2}

的解:

真

a^{2} - b^{2} = \frac{1}{2}

代入

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = - \frac{2 2 - 1}{2}

- \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}^{2} - (- \frac{2 2 - 1}{2})^{2} = \frac{1}{2}

整理后得

\frac{1}{2} = \frac{1}{2}

真

检验

a = \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

的解:

真

a^{2} - b^{2} = \frac{1}{2}

代入

a = \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

(\frac{7}{2 2 2 - 1})^{2} - (\frac{2 2 - 1}{2})^{2} = \frac{1}{2}

整理后得

\frac{1}{2} = \frac{1}{2}

真

将它们代入

2 ab = \frac{7}{2}

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = - \frac{2 2 - 1}{2}

的解:

真

2 ab = \frac{7}{2}

代入

a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = - \frac{2 2 - 1}{2}

2 - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} (- \frac{2 2 - 1}{2}) = \frac{7}{2}

整理后得

\frac{7}{2} = \frac{7}{2}

真

检验

a = \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

的解:

真

2 ab = \frac{7}{2}

代入

a = \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

2 \cdot \frac{7}{2 2 2 - 1} \cdot \frac{2 2 - 1}{2} = \frac{7}{2}

整理后得

\frac{7}{2} = \frac{7}{2}

真

因而，

a^{2} - b^{2} = \frac{1}{2}, 2 ab = \frac{7}{2}

最后的解是

a = \frac{7}{2 2 2 - 1}, a = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}}, b = \frac{2 2 - 1}{2} b = - \frac{2 2 - 1}{2}

u = a + bi

代回

u = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i

解

u^{2} = \frac{1}{2} - i \frac{7}{2} : u = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

u^{2} = \frac{1}{2} - i \frac{7}{2}

替代

u = a + bi

(a + bi)^{2} = \frac{1}{2} - i \frac{7}{2}

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = \frac{1}{2} - i \frac{7}{2}

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{7}{2}]

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{7}{2}] : a = - \frac{7}{2 2 2 - 1}, a = - \frac{7}{2 ( - 2 2 - 1 )}, b = \frac{2 2 - 1}{2} b = - \frac{2 2 - 1}{2}

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{7}{2}]

对于

2 ab = - \frac{7}{2}

将

a

移到一边:

a = - \frac{7}{4 b}

2 ab = - \frac{7}{2}

两边除以

2 b

2 ab = - \frac{7}{2}

两边除以

2 b

\frac{2 ab}{2 b} = \frac{- \frac{7}{2}}{2 b}

化简

\frac{2 ab}{2 b} = \frac{- \frac{7}{2}}{2 b}

化简

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

数字相除:

\frac{2}{2} = 1

= \frac{ab}{b}

约分:

b

= a

化简

\frac{- \frac{7}{2}}{2 b} : - \frac{7}{4 b}

\frac{- \frac{7}{2}}{2 b}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{7}{2}}{2 b}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{7}{2}}{2 b} = \frac{7}{2 \cdot 2 b}

= - \frac{7}{2 \cdot 2 b}

数字相乘:

2 \cdot 2 = 4

= - \frac{7}{4 b}

a = - \frac{7}{4 b}

a = - \frac{7}{4 b}

a = - \frac{7}{4 b}

将解

a = - \frac{7}{4 b}

代入

a^{2} - b^{2} = \frac{1}{2}

对于

a^{2} - b^{2} = \frac{1}{2}

，用

- \frac{7}{4 b}

替代

a : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

对于

a^{2} - b^{2} = \frac{1}{2}

，用

- \frac{7}{4 b}

替代

a

(- \frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

解

(- \frac{7}{4 b})^{2} - b^{2} = \frac{1}{2} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

(- \frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

乘以最小公倍数

(- \frac{7}{4 b})^{2} - b^{2} = \frac{1}{2}

化简

(- \frac{7}{4 b})^{2} : \frac{7}{16 b ^{2}}

(- \frac{7}{4 b})^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- \frac{7}{4 b})^{2} = (\frac{7}{4 b})^{2}

= (\frac{7}{4 b})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 7 ) ^{2}}{( 4 b ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(4 b)^{2} = 4^{2} b^{2}

= \frac{( 7 ) ^{2}}{4 ^{2} b ^{2}}

(7)^{2} : 7

使用根式运算法则:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 7^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 7

= \frac{7}{4 ^{2} b ^{2}}

4^{2} = 16

= \frac{7}{16 b ^{2}}

\frac{7}{16 b ^{2}} - b^{2} = \frac{1}{2}

找到

16 b^{2}, 2

的最小公倍数:

16 b^{2}

16 b^{2}, 2

最小公倍数 (LCM)

16, 2

的最小公倍数:

16

16, 2

最小公倍数 (LCM)

16

质因数分解:

2 \cdot 2 \cdot 2 \cdot 2

16

16

除以

216 = 8 \cdot 2

= 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

16

或

2

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

计算出由出现在

16 b^{2}

或

2

中的因子组成的表达式

= 16 b^{2}

乘以最小公倍数=

16 b^{2}

\frac{7}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{7}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{7}{16 b ^{2}} \cdot 16 b^{2} : 7

\frac{7}{16 b ^{2}} \cdot 16 b^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{7 \cdot 16 b ^{2}}{16 b ^{2}}

约分:

16

= \frac{7 b ^{2}}{b ^{2}}

约分:

b^{2}

= 7

化简

- b^{2} \cdot 16 b^{2} : - 16 b^{4}

- b^{2} \cdot 16 b^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 16 b^{4}

化简

\frac{1}{2} \cdot 16 b^{2} : 8 b^{2}

\frac{1}{2} \cdot 16 b^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 16}{2} b^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

数字相乘:

1 \cdot 16 = 16

= \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= 8

= 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

7 - 16 b^{4} = 8 b^{2}

解

7 - 16 b^{4} = 8 b^{2} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

7 - 16 b^{4} = 8 b^{2}

将

8 b^{2}

para o lado esquerdo

7 - 16 b^{4} = 8 b^{2}

两边减去

8 b^{2}

7 - 16 b^{4} - 8 b^{2} = 8 b^{2} - 8 b^{2}

化简

7 - 16 b^{4} - 8 b^{2} = 0

7 - 16 b^{4} - 8 b^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 b^{4} - 8 b^{2} + 7 = 0

用

u = b^{2}

和

u^{2} = b^{4}

改写方程式

- 16 u^{2} - 8 u + 7 = 0

解

- 16 u^{2} - 8 u + 7 = 0 : u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

- 16 u^{2} - 8 u + 7 = 0

使用求根公式求解

- 16 u^{2} - 8 u + 7 = 0

二次方程求根公式:

若

a = - 16, b = - 8, c = 7

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 7}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 7}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 7 = 162

(- 8)^{2} - 4 (- 16) \cdot 7

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 7

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 7

数字相乘:

4 \cdot 16 \cdot 7 = 448

= 8^{2} + 448

8^{2} = 64

= 64 + 448

数字相加:

64 + 448 = 512

= 512

512

质因数分解:

2^{9}

512

512

除以

2512 = 256 \cdot 2

= 2 \cdot 256

256

除以

2256 = 128 \cdot 2

= 2 \cdot 2 \cdot 128

128

除以

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 64

64

除以

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{9}

= 2^{9}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{8} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{8}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{8} = 2^{\frac{8}{2}} = 2^{4}

= 2^{4} 2

整理后得

= 162

u_{1, 2} = \frac{- ( - 8 ) \pm 16 2}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 16 2}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16 2}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16 2}{2 ( - 16 )} : - \frac{1 + 2 2}{4}

\frac{- ( - 8 ) + 16 2}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 + 16 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 + 16 2}{- 32}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8 + 16 2}{32}

消掉

\frac{8 + 16 2}{32} : \frac{1 + 2 2}{4}

\frac{8 + 16 2}{32}

分解

8 + 162 : 8 (1 + 22)

8 + 162

改写为

= 8 \cdot 1 + 8 \cdot 22

因式分解出通项

8

= 8 (1 + 22)

= \frac{8 ( 1 + 2 2 )}{32}

约分:

8

= \frac{1 + 2 2}{4}

= - \frac{1 + 2 2}{4}

u = \frac{- ( - 8 ) - 16 2}{2 ( - 16 )} : \frac{2 2 - 1}{4}

\frac{- ( - 8 ) - 16 2}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 - 16 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 - 16 2}{- 32}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

8 - 162 = - (162 - 8)

= \frac{16 2 - 8}{32}

分解

162 - 8 : 8 (22 - 1)

162 - 8

改写为

= 8 \cdot 22 - 8 \cdot 1

因式分解出通项

8

= 8 (22 - 1)

= \frac{8 ( 2 2 - 1 )}{32}

约分:

8

= \frac{2 2 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

u = - \frac{1 + 2 2}{4}, u = \frac{2 2 - 1}{4}

代回

u = b^{2}

，求解

b

解

b^{2} = - \frac{1 + 2 2}{4} : b \in R

无解

b^{2} = - \frac{1 + 2 2}{4}

x^{2}

在

x

内不能为负

\in R

b \in R 无解

解

b^{2} = \frac{2 2 - 1}{4} : b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

b^{2} = \frac{2 2 - 1}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

b = \frac{2 2 - 1}{4}, b = - \frac{2 2 - 1}{4}

\frac{2 2 - 1}{4} = \frac{2 2 - 1}{2}

\frac{2 2 - 1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{2 2 - 1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{2 2 - 1}{2}

- \frac{2 2 - 1}{4} = - \frac{2 2 - 1}{2}

- \frac{2 2 - 1}{4}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{2 2 - 1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{2 2 - 1}{2}

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

解为

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(- \frac{7}{4 b})^{2} - b^{2}

的分母，令其等于零

解

4 b = 0 : b = 0

4 b = 0

两边除以

4

4 b = 0

两边除以

4

\frac{4 b}{4} = \frac{0}{4}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

将解

b = \frac{2 2 - 1}{2}, b = - \frac{2 2 - 1}{2}

代入

2 ab = - \frac{7}{2}

对于

2 ab = - \frac{7}{2}

，用

\frac{2 2 - 1}{2}

替代

b : a = - \frac{7}{2 2 2 - 1}

对于

2 ab = - \frac{7}{2}

，用

\frac{2 2 - 1}{2}

替代

b

2 a \frac{2 2 - 1}{2} = - \frac{7}{2}

解

2 a \frac{2 2 - 1}{2} = - \frac{7}{2} : a = - \frac{7}{2 2 2 - 1}

2 a \frac{2 2 - 1}{2} = - \frac{7}{2}

在两边乘以

2

2 a \frac{2 2 - 1}{2} = - \frac{7}{2}

在两边乘以

2

2 \cdot 2 a \frac{2 2 - 1}{2} = 2 (- \frac{7}{2})

化简

2 \cdot 2 a \frac{2 2 - 1}{2} = 2 (- \frac{7}{2})

化简

2 \cdot 2 a \frac{2 2 - 1}{2} : 2 a 22 - 1

2 \cdot 2 a \frac{2 2 - 1}{2}

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= 2^{2} a \frac{2 2 - 1}{2}

使用分式法则:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 ^{2} a 2 2 - 1}{2}

消掉

\frac{2 ^{2} a 2 2 - 1}{2} : 2 a 22 - 1

\frac{2 ^{2} a 2 2 - 1}{2}

\frac{2 ^{2}}{2} = 2

\frac{2 ^{2}}{2}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

2^{2} = 2 \cdot 2

= \frac{2 \cdot 2}{2}

约分:

2

= 2

= 2 a 22 - 1

= 2 a 22 - 1

化简

2 (- \frac{7}{2}) : - 7

2 (- \frac{7}{2})

使用法则:

a (- b) = - ab

2 (- \frac{7}{2}) = - 2 \cdot \frac{7}{2}

= - 2 \cdot \frac{7}{2}

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{2}{1} \cdot \frac{7}{2}

交叉约去公约数:

2

= - \frac{7}{1}

使用分式法则:

\frac{a}{1} = a

= - 7

2 a 22 - 1 = - 7

2 a 22 - 1 = - 7

2 a 22 - 1 = - 7

两边除以

2 22 - 1

2 a 22 - 1 = - 7

两边除以

2 22 - 1

\frac{2 a 2 2 - 1}{2 2 2 - 1} = \frac{- 7}{2 2 2 - 1}

化简

a = - \frac{7}{2 2 2 - 1}

a = - \frac{7}{2 2 2 - 1}

对于

2 ab = - \frac{7}{2}

，用

- \frac{2 2 - 1}{2}

替代

b : a = - \frac{7}{2 ( - 2 2 - 1 )}

对于

2 ab = - \frac{7}{2}

，用

- \frac{2 2 - 1}{2}

替代

b

2 a (- \frac{2 2 - 1}{2}) = - \frac{7}{2}

解

2 a (- \frac{2 2 - 1}{2}) = - \frac{7}{2} : a = - \frac{7}{2 ( - 2 2 - 1 )}

2 a (- \frac{2 2 - 1}{2}) = - \frac{7}{2}

两边除以

2 (- \frac{2 2 - 1}{2})

2 a (- \frac{2 2 - 1}{2}) = - \frac{7}{2}

两边除以

2 (- \frac{2 2 - 1}{2})

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} = \frac{- \frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} = \frac{- \frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} : a

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )}

化简

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )} : \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

\frac{2 a ( - \frac{2 2 - 1}{2} )}{2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 a (- \frac{2 2 - 1}{2}) = - 2 a \frac{2 2 - 1}{2}

= \frac{- 2 a \frac{2 2 - 1}{2}}{2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{2 2 - 1}{2}) = - 2 \cdot \frac{2 2 - 1}{2}

= \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

= \frac{- 2 a \frac{2 2 - 1}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

约分:

- 2

= \frac{a \frac{2 2 - 1}{2}}{\frac{2 2 - 1}{2}}

约分:

\frac{2 2 - 1}{2}

= a

化简

\frac{- \frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )} : - \frac{7}{2 ( - 2 2 - 1 )}

\frac{- \frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{7}{2}}{2 ( - \frac{2 2 - 1}{2} )}

使用法则:

a (- b) = - ab

2 (- \frac{2 2 - 1}{2}) = - 2 \cdot \frac{2 2 - 1}{2}

= - \frac{\frac{7}{2}}{- 2 \cdot \frac{2 2 - 1}{2}}

- 2 \cdot \frac{2 2 - 1}{2} = - 22 - 1

- 2 \cdot \frac{2 2 - 1}{2}

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{2}{1} \cdot \frac{2 2 - 1}{2}

交叉约去公约数:

2

= - \frac{2 2 - 1}{1}

使用分式法则:

\frac{a}{1} = a

\frac{2 2 - 1}{1} = 22 - 1

= - 22 - 1

= - \frac{\frac{7}{2}}{- 2 2 - 1}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

\frac{\frac{7}{2}}{- 2 2 - 1} = \frac{7}{2 ( - 2 2 - 1 )}

= - \frac{7}{2 ( - 2 2 - 1 )}

a = - \frac{7}{2 ( - 2 2 - 1 )}

a = - \frac{7}{2 ( - 2 2 - 1 )}

a = - \frac{7}{2 ( - 2 2 - 1 )}

将解代入原方程进行验证

将它们代入

a^{2} - b^{2} = \frac{1}{2}

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{7}{2 ( - 2 2 - 1 )}, b = - \frac{2 2 - 1}{2}

的解:

真

a^{2} - b^{2} = \frac{1}{2}

代入

a = - \frac{7}{2 ( - 2 2 - 1 )}, b = - \frac{2 2 - 1}{2}

- \frac{7}{2 ( - 2 2 - 1 )}^{2} - (- \frac{2 2 - 1}{2})^{2} = \frac{1}{2}

整理后得

\frac{1}{2} = \frac{1}{2}

真

检验

a = - \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

的解:

真

a^{2} - b^{2} = \frac{1}{2}

代入

a = - \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

(- \frac{7}{2 2 2 - 1})^{2} - (\frac{2 2 - 1}{2})^{2} = \frac{1}{2}

整理后得

\frac{1}{2} = \frac{1}{2}

真

将它们代入

2 ab = - \frac{7}{2}

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{7}{2 ( - 2 2 - 1 )}, b = - \frac{2 2 - 1}{2}

的解:

真

2 ab = - \frac{7}{2}

代入

a = - \frac{7}{2 ( - 2 2 - 1 )}, b = - \frac{2 2 - 1}{2}

2 - \frac{7}{2 ( - 2 2 - 1 )} (- \frac{2 2 - 1}{2}) = - \frac{7}{2}

整理后得

- \frac{7}{2} = - \frac{7}{2}

真

检验

a = - \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

的解:

真

2 ab = - \frac{7}{2}

代入

a = - \frac{7}{2 2 2 - 1}, b = \frac{2 2 - 1}{2}

2 (- \frac{7}{2 2 2 - 1}) \frac{2 2 - 1}{2} = - \frac{7}{2}

整理后得

- \frac{7}{2} = - \frac{7}{2}

真

因而，

a^{2} - b^{2} = \frac{1}{2}, 2 ab = - \frac{7}{2}

最后的解是

a = - \frac{7}{2 2 2 - 1}, a = - \frac{7}{2 ( - 2 2 - 1 )}, b = \frac{2 2 - 1}{2} b = - \frac{2 2 - 1}{2}

u = a + bi

代回

u = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

解为

u = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i, u = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, u = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

u = cos (x)

代回

cos (x) = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

cos (x) = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i, cos (x) = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

cos (x) = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i :

无解

cos (x) = \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i

化简

\frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i : \frac{7 ( - 1 + 2 2 + 2 - 2 + 4 2 )}{14} + i \frac{- 1 + 2 2}{2}

\frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i

\frac{7}{2 2 2 - 1} = \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

\frac{7}{2 2 2 - 1}

乘以共轭根式

\frac{2 2 - 1}{2 2 - 1}

= \frac{7 2 2 - 1}{2 2 2 - 1 2 2 - 1}

2 22 - 1 22 - 1 = 42 - 2

2 22 - 1 22 - 1

使用根式运算法则:

a a = a

22 - 1 22 - 1 = 22 - 1

= 2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= \frac{7 2 2 - 1}{4 2 - 2}

乘以共轭根式

\frac{4 2 + 2}{4 2 + 2}

= \frac{7 2 2 - 1 ( 4 2 + 2 )}{( 4 2 - 2 ) ( 4 2 + 2 )}

(42 - 2) (42 + 2) = 28

(42 - 2) (42 + 2)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = 42, b = 2

= (42)^{2} - 2^{2}

化简

(42)^{2} - 2^{2} : 28

(42)^{2} - 2^{2}

(42)^{2} = 32

(42)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= 32 - 4

数字相减:

32 - 4 = 28

= 28

= 28

= \frac{7 ( 4 2 + 2 ) 2 2 - 1}{28}

分解

42 + 2 : 2 (22 + 1)

42 + 2

改写为

= 2 \cdot 22 + 2 \cdot 1

因式分解出通项

2

= 2 (22 + 1)

= \frac{7 \cdot 2 ( 2 2 + 1 ) 2 2 - 1}{28}

约分:

2

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + i \frac{2 2 - 1}{2}

将

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + \frac{2 2 - 1}{2} i

改写成标准复数形式:

\frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} = \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

分解

14 : 2 \cdot 7

因式分解

14 = 2 \cdot 7

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

消掉

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7} : \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用根式运算法则:

n a = a^{\frac{1}{n}}

7 = 7^{\frac{1}{2}}

= \frac{7 ^{\frac{1}{2}} ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{7 ^{\frac{1}{2}}}{7 ^{1}} = \frac{1}{7 ^{1 - \frac{1}{2}}}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{- \frac{1}{2} + 1}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

7^{\frac{1}{2}} = 7

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

乘开

(22 + 1) 22 - 1 : 2 42 - 2 + 22 - 1

(22 + 1) 22 - 1

使用分配律:

a (b + c) = ab + a c

a = 22 - 1, b = 22, c = 1

= 22 - 1 \cdot 22 + 22 - 1 \cdot 1

= 22 22 - 1 + 1 \cdot 22 - 1

化简

22 22 - 1 + 1 \cdot 22 - 1 : 2 42 - 2 + 22 - 1

22 22 - 1 + 1 \cdot 22 - 1

22 22 - 1 = 2 42 - 2

22 22 - 1

使用根式运算法则:

a b = a \cdot b

2 22 - 1 = 2 (22 - 1)

= 2 2 (22 - 1)

乘开

2 (22 - 1) : 42 - 2

2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= 2 42 - 2

1 \cdot 22 - 1 = 22 - 1

1 \cdot 22 - 1

乘以:

1 \cdot 22 - 1 = 22 - 1

= 22 - 1

= 2 42 - 2 + 22 - 1

= 2 42 - 2 + 22 - 1

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 2 - 1}{2} i = \frac{i 2 2 - 1}{2}

\frac{2 2 - 1}{2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 2 - 1 i}{2}

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7} = \frac{2 4 2 - 2}{2 7} + \frac{2 2 - 1}{2 7}

= \frac{2 4 2 - 2}{2 7} + \frac{2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

\frac{2 4 2 - 2}{2 7} = \frac{4 2 - 2}{7}

\frac{2 4 2 - 2}{2 7}

数字相除:

\frac{2}{2} = 1

= \frac{4 2 - 2}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{4 2 - 2}{7}

= \frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

将复数的实部和虚部分组

= \frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} + \frac{2 2 - 1}{2} i

\frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} = \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

\frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7}

将项转换为分式:

\frac{4 2 - 2}{7} = \frac{\frac{4 \cdot 2 - 2}{7} 2 7}{2 7}

= \frac{\frac{4 2 - 2}{7} \cdot 2 7}{2 7} + \frac{2 2 - 1}{2 7}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{\frac{4 2 - 2}{7} \cdot 2 7 + 2 2 - 1}{2 7}

\frac{4 2 - 2}{7} \cdot 27 = 2 42 - 2

\frac{4 2 - 2}{7} \cdot 27

使用根式运算法则:

a b = a \cdot b

7 \frac{4 2 - 2}{7} = 7 \cdot \frac{4 2 - 2}{7}

= 2 7 \cdot \frac{4 2 - 2}{7}

\frac{4 2 - 2}{7} \cdot 7 = 42 - 2

\frac{4 2 - 2}{7} \cdot 7

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 4 2 - 2 ) \cdot 7}{7}

约分:

7

= 42 - 2

= 2 42 - 2

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7}

有理化:

\frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7}

乘以共轭根式

\frac{7}{7}

= \frac{( 4 2 - 2 \cdot 2 + 2 2 - 1 ) 7}{2 7 7}

277 = 14

277

使用根式运算法则:

a a = a

77 = 7

= 2 \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

无解

cos (x) = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i :

无解

cos (x) = - \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i

化简

- \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i : \frac{7 ( - - 1 + 2 2 - 2 - 2 + 4 2 )}{14} - i \frac{- 1 + 2 2}{2}

- \frac{7}{2 ^{2} \cdot \frac{2 2 - 1}{2}} - \frac{2 2 - 1}{2} i

乘

2^{2} \cdot \frac{2 2 - 1}{2} : 2 22 - 1

2^{2} \cdot \frac{2 2 - 1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 2 - 1 \cdot 2 ^{2}}{2}

约分:

2

= 2 22 - 1

= - \frac{7}{2 2 2 - 1} - i \frac{2 2 - 1}{2}

\frac{7}{2 2 2 - 1} = \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

\frac{7}{2 2 2 - 1}

乘以共轭根式

\frac{2 2 - 1}{2 2 - 1}

= \frac{7 2 2 - 1}{2 2 2 - 1 2 2 - 1}

2 22 - 1 22 - 1 = 42 - 2

2 22 - 1 22 - 1

使用根式运算法则:

a a = a

22 - 1 22 - 1 = 22 - 1

= 2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= \frac{7 2 2 - 1}{4 2 - 2}

乘以共轭根式

\frac{4 2 + 2}{4 2 + 2}

= \frac{7 2 2 - 1 ( 4 2 + 2 )}{( 4 2 - 2 ) ( 4 2 + 2 )}

(42 - 2) (42 + 2) = 28

(42 - 2) (42 + 2)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = 42, b = 2

= (42)^{2} - 2^{2}

化简

(42)^{2} - 2^{2} : 28

(42)^{2} - 2^{2}

(42)^{2} = 32

(42)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= 32 - 4

数字相减:

32 - 4 = 28

= 28

= 28

= \frac{7 ( 4 2 + 2 ) 2 2 - 1}{28}

分解

42 + 2 : 2 (22 + 1)

42 + 2

改写为

= 2 \cdot 22 + 2 \cdot 1

因式分解出通项

2

= 2 (22 + 1)

= \frac{7 \cdot 2 ( 2 2 + 1 ) 2 2 - 1}{28}

约分:

2

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

= - \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - i \frac{2 2 - 1}{2}

将

- \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - \frac{2 2 - 1}{2} i

改写成标准复数形式:

\frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

- \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} = \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

分解

14 : 2 \cdot 7

因式分解

14 = 2 \cdot 7

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

消掉

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7} : \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用根式运算法则:

n a = a^{\frac{1}{n}}

7 = 7^{\frac{1}{2}}

= \frac{7 ^{\frac{1}{2}} ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{7 ^{\frac{1}{2}}}{7 ^{1}} = \frac{1}{7 ^{1 - \frac{1}{2}}}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{- \frac{1}{2} + 1}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

7^{\frac{1}{2}} = 7

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

乘开

(22 + 1) 22 - 1 : 2 42 - 2 + 22 - 1

(22 + 1) 22 - 1

使用分配律:

a (b + c) = ab + a c

a = 22 - 1, b = 22, c = 1

= 22 - 1 \cdot 22 + 22 - 1 \cdot 1

= 22 22 - 1 + 1 \cdot 22 - 1

化简

22 22 - 1 + 1 \cdot 22 - 1 : 2 42 - 2 + 22 - 1

22 22 - 1 + 1 \cdot 22 - 1

22 22 - 1 = 2 42 - 2

22 22 - 1

使用根式运算法则:

a b = a \cdot b

2 22 - 1 = 2 (22 - 1)

= 2 2 (22 - 1)

乘开

2 (22 - 1) : 42 - 2

2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= 2 42 - 2

1 \cdot 22 - 1 = 22 - 1

1 \cdot 22 - 1

乘以:

1 \cdot 22 - 1 = 22 - 1

= 22 - 1

= 2 42 - 2 + 22 - 1

= 2 42 - 2 + 22 - 1

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 2 - 1}{2} i = \frac{i 2 2 - 1}{2}

\frac{2 2 - 1}{2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 2 - 1 i}{2}

= - \frac{2 4 2 - 2 + 2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7} = - (\frac{2 4 2 - 2}{2 7}) - (\frac{2 2 - 1}{2 7})

= - (\frac{2 4 2 - 2}{2 7}) - (\frac{2 2 - 1}{2 7}) - \frac{i 2 2 - 1}{2}

去除括号:

(a) = a

= - \frac{2 4 2 - 2}{2 7} - \frac{2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

\frac{2 4 2 - 2}{2 7} = \frac{4 2 - 2}{7}

\frac{2 4 2 - 2}{2 7}

数字相除:

\frac{2}{2} = 1

= \frac{4 2 - 2}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{4 2 - 2}{7}

= - \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

将复数的实部和虚部分组

= - \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} - \frac{2 2 - 1}{2} i

- \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} = \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

- \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7}

将项转换为分式:

\frac{4 2 - 2}{7} = \frac{\frac{4 \cdot 2 - 2}{7} 2 7}{2 7}

= - \frac{\frac{4 2 - 2}{7} \cdot 2 7}{2 7} - \frac{2 2 - 1}{2 7}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- \frac{4 2 - 2}{7} \cdot 2 7 - 2 2 - 1}{2 7}

\frac{4 2 - 2}{7} \cdot 27 = 2 42 - 2

\frac{4 2 - 2}{7} \cdot 27

使用根式运算法则:

a b = a \cdot b

7 \frac{4 2 - 2}{7} = 7 \cdot \frac{4 2 - 2}{7}

= 2 7 \cdot \frac{4 2 - 2}{7}

\frac{4 2 - 2}{7} \cdot 7 = 42 - 2

\frac{4 2 - 2}{7} \cdot 7

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 4 2 - 2 ) \cdot 7}{7}

约分:

7

= 42 - 2

= 2 42 - 2

= \frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

\frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

有理化:

\frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

\frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

乘以共轭根式

\frac{7}{7}

= \frac{( - 4 2 - 2 \cdot 2 - 2 2 - 1 ) 7}{2 7 7}

277 = 14

277

使用根式运算法则:

a a = a

77 = 7

= 2 \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

无解

cos (x) = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i :

无解

cos (x) = - \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i

化简

- \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i : \frac{7 ( - - 1 + 2 2 - 2 - 2 + 4 2 )}{14} + i \frac{- 1 + 2 2}{2}

- \frac{7}{2 2 2 - 1} + \frac{2 2 - 1}{2} i

\frac{7}{2 2 2 - 1} = \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

\frac{7}{2 2 2 - 1}

乘以共轭根式

\frac{2 2 - 1}{2 2 - 1}

= \frac{7 2 2 - 1}{2 2 2 - 1 2 2 - 1}

2 22 - 1 22 - 1 = 42 - 2

2 22 - 1 22 - 1

使用根式运算法则:

a a = a

22 - 1 22 - 1 = 22 - 1

= 2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= \frac{7 2 2 - 1}{4 2 - 2}

乘以共轭根式

\frac{4 2 + 2}{4 2 + 2}

= \frac{7 2 2 - 1 ( 4 2 + 2 )}{( 4 2 - 2 ) ( 4 2 + 2 )}

(42 - 2) (42 + 2) = 28

(42 - 2) (42 + 2)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = 42, b = 2

= (42)^{2} - 2^{2}

化简

(42)^{2} - 2^{2} : 28

(42)^{2} - 2^{2}

(42)^{2} = 32

(42)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= 32 - 4

数字相减:

32 - 4 = 28

= 28

= 28

= \frac{7 ( 4 2 + 2 ) 2 2 - 1}{28}

分解

42 + 2 : 2 (22 + 1)

42 + 2

改写为

= 2 \cdot 22 + 2 \cdot 1

因式分解出通项

2

= 2 (22 + 1)

= \frac{7 \cdot 2 ( 2 2 + 1 ) 2 2 - 1}{28}

约分:

2

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

= - \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + i \frac{2 2 - 1}{2}

将

- \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + \frac{2 2 - 1}{2} i

改写成标准复数形式:

\frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

- \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} + \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} = \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

分解

14 : 2 \cdot 7

因式分解

14 = 2 \cdot 7

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

消掉

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7} : \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用根式运算法则:

n a = a^{\frac{1}{n}}

7 = 7^{\frac{1}{2}}

= \frac{7 ^{\frac{1}{2}} ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{7 ^{\frac{1}{2}}}{7 ^{1}} = \frac{1}{7 ^{1 - \frac{1}{2}}}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{- \frac{1}{2} + 1}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

7^{\frac{1}{2}} = 7

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

乘开

(22 + 1) 22 - 1 : 2 42 - 2 + 22 - 1

(22 + 1) 22 - 1

使用分配律:

a (b + c) = ab + a c

a = 22 - 1, b = 22, c = 1

= 22 - 1 \cdot 22 + 22 - 1 \cdot 1

= 22 22 - 1 + 1 \cdot 22 - 1

化简

22 22 - 1 + 1 \cdot 22 - 1 : 2 42 - 2 + 22 - 1

22 22 - 1 + 1 \cdot 22 - 1

22 22 - 1 = 2 42 - 2

22 22 - 1

使用根式运算法则:

a b = a \cdot b

2 22 - 1 = 2 (22 - 1)

= 2 2 (22 - 1)

乘开

2 (22 - 1) : 42 - 2

2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= 2 42 - 2

1 \cdot 22 - 1 = 22 - 1

1 \cdot 22 - 1

乘以:

1 \cdot 22 - 1 = 22 - 1

= 22 - 1

= 2 42 - 2 + 22 - 1

= 2 42 - 2 + 22 - 1

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 2 - 1}{2} i = \frac{i 2 2 - 1}{2}

\frac{2 2 - 1}{2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 2 - 1 i}{2}

= - \frac{2 4 2 - 2 + 2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7} = - (\frac{2 4 2 - 2}{2 7}) - (\frac{2 2 - 1}{2 7})

= - (\frac{2 4 2 - 2}{2 7}) - (\frac{2 2 - 1}{2 7}) + \frac{i 2 2 - 1}{2}

去除括号:

(a) = a

= - \frac{2 4 2 - 2}{2 7} - \frac{2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

\frac{2 4 2 - 2}{2 7} = \frac{4 2 - 2}{7}

\frac{2 4 2 - 2}{2 7}

数字相除:

\frac{2}{2} = 1

= \frac{4 2 - 2}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{4 2 - 2}{7}

= - \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} + \frac{i 2 2 - 1}{2}

将复数的实部和虚部分组

= - \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} + \frac{2 2 - 1}{2} i

- \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7} = \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

- \frac{4 2 - 2}{7} - \frac{2 2 - 1}{2 7}

将项转换为分式:

\frac{4 2 - 2}{7} = \frac{\frac{4 \cdot 2 - 2}{7} 2 7}{2 7}

= - \frac{\frac{4 2 - 2}{7} \cdot 2 7}{2 7} - \frac{2 2 - 1}{2 7}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- \frac{4 2 - 2}{7} \cdot 2 7 - 2 2 - 1}{2 7}

\frac{4 2 - 2}{7} \cdot 27 = 2 42 - 2

\frac{4 2 - 2}{7} \cdot 27

使用根式运算法则:

a b = a \cdot b

7 \frac{4 2 - 2}{7} = 7 \cdot \frac{4 2 - 2}{7}

= 2 7 \cdot \frac{4 2 - 2}{7}

\frac{4 2 - 2}{7} \cdot 7 = 42 - 2

\frac{4 2 - 2}{7} \cdot 7

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 4 2 - 2 ) \cdot 7}{7}

约分:

7

= 42 - 2

= 2 42 - 2

= \frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

\frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

有理化:

\frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

\frac{- 2 4 2 - 2 - 2 2 - 1}{2 7}

乘以共轭根式

\frac{7}{7}

= \frac{( - 4 2 - 2 \cdot 2 - 2 2 - 1 ) 7}{2 7 7}

277 = 14

277

使用根式运算法则:

a a = a

77 = 7

= 2 \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14}

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

= \frac{7 ( - 2 4 2 - 2 - 2 2 - 1 )}{14} + \frac{2 2 - 1}{2} i

无解

cos (x) = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i :

无解

cos (x) = - \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

化简

- \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i : \frac{7 ( - 1 + 2 2 + 2 - 2 + 4 2 )}{14} - i \frac{- 1 + 2 2}{2}

- \frac{7}{2 ( - 2 2 - 1 )} - \frac{2 2 - 1}{2} i

去除括号:

(- a) = - a

= - \frac{7}{- 2 2 2 - 1} - \frac{2 2 - 1}{2} i

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - (- \frac{7}{2 2 2 - 1}) - i \frac{2 2 - 1}{2}

使用法则

- (- a) = a

= \frac{7}{2 2 2 - 1} - \frac{2 2 - 1}{2} i

\frac{7}{2 2 2 - 1} = \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

\frac{7}{2 2 2 - 1}

乘以共轭根式

\frac{2 2 - 1}{2 2 - 1}

= \frac{7 2 2 - 1}{2 2 2 - 1 2 2 - 1}

2 22 - 1 22 - 1 = 42 - 2

2 22 - 1 22 - 1

使用根式运算法则:

a a = a

22 - 1 22 - 1 = 22 - 1

= 2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= \frac{7 2 2 - 1}{4 2 - 2}

乘以共轭根式

\frac{4 2 + 2}{4 2 + 2}

= \frac{7 2 2 - 1 ( 4 2 + 2 )}{( 4 2 - 2 ) ( 4 2 + 2 )}

(42 - 2) (42 + 2) = 28

(42 - 2) (42 + 2)

使用平方差公式:

(a - b) (a + b) = a^{2} - b^{2}

a = 42, b = 2

= (42)^{2} - 2^{2}

化简

(42)^{2} - 2^{2} : 28

(42)^{2} - 2^{2}

(42)^{2} = 32

(42)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

2^{2} = 4

2^{2}

2^{2} = 4

= 4

= 32 - 4

数字相减:

32 - 4 = 28

= 28

= 28

= \frac{7 ( 4 2 + 2 ) 2 2 - 1}{28}

分解

42 + 2 : 2 (22 + 1)

42 + 2

改写为

= 2 \cdot 22 + 2 \cdot 1

因式分解出通项

2

= 2 (22 + 1)

= \frac{7 \cdot 2 ( 2 2 + 1 ) 2 2 - 1}{28}

约分:

2

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - i \frac{2 2 - 1}{2}

将

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - \frac{2 2 - 1}{2} i

改写成标准复数形式:

\frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} - \frac{2 2 - 1}{2} i

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14} = \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{14}

分解

14 : 2 \cdot 7

因式分解

14 = 2 \cdot 7

= \frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

消掉

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7} : \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

\frac{7 ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用根式运算法则:

n a = a^{\frac{1}{n}}

7 = 7^{\frac{1}{2}}

= \frac{7 ^{\frac{1}{2}} ( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{7 ^{\frac{1}{2}}}{7 ^{1}} = \frac{1}{7 ^{1 - \frac{1}{2}}}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{- \frac{1}{2} + 1}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 \cdot 7 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

7^{\frac{1}{2}} = 7

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

= \frac{( 2 2 + 1 ) 2 2 - 1}{2 7}

乘开

(22 + 1) 22 - 1 : 2 42 - 2 + 22 - 1

(22 + 1) 22 - 1

使用分配律:

a (b + c) = ab + a c

a = 22 - 1, b = 22, c = 1

= 22 - 1 \cdot 22 + 22 - 1 \cdot 1

= 22 22 - 1 + 1 \cdot 22 - 1

化简

22 22 - 1 + 1 \cdot 22 - 1 : 2 42 - 2 + 22 - 1

22 22 - 1 + 1 \cdot 22 - 1

22 22 - 1 = 2 42 - 2

22 22 - 1

使用根式运算法则:

a b = a \cdot b

2 22 - 1 = 2 (22 - 1)

= 2 2 (22 - 1)

乘开

2 (22 - 1) : 42 - 2

2 (22 - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 22, c = 1

= 2 \cdot 22 - 2 \cdot 1

化简

2 \cdot 22 - 2 \cdot 1 : 42 - 2

2 \cdot 22 - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

= 42 - 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 42 - 2

= 42 - 2

= 2 42 - 2

1 \cdot 22 - 1 = 22 - 1

1 \cdot 22 - 1

乘以:

1 \cdot 22 - 1 = 22 - 1

= 22 - 1

= 2 42 - 2 + 22 - 1

= 2 42 - 2 + 22 - 1

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 2 - 1}{2} i = \frac{i 2 2 - 1}{2}

\frac{2 2 - 1}{2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 2 - 1 i}{2}

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7} = \frac{2 4 2 - 2}{2 7} + \frac{2 2 - 1}{2 7}

= \frac{2 4 2 - 2}{2 7} + \frac{2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

\frac{2 4 2 - 2}{2 7} = \frac{4 2 - 2}{7}

\frac{2 4 2 - 2}{2 7}

数字相除:

\frac{2}{2} = 1

= \frac{4 2 - 2}{7}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{4 2 - 2}{7}

= \frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} - \frac{i 2 2 - 1}{2}

将复数的实部和虚部分组

= \frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} - \frac{2 2 - 1}{2} i

\frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7} = \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

\frac{4 2 - 2}{7} + \frac{2 2 - 1}{2 7}

将项转换为分式:

\frac{4 2 - 2}{7} = \frac{\frac{4 \cdot 2 - 2}{7} 2 7}{2 7}

= \frac{\frac{4 2 - 2}{7} \cdot 2 7}{2 7} + \frac{2 2 - 1}{2 7}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{\frac{4 2 - 2}{7} \cdot 2 7 + 2 2 - 1}{2 7}

\frac{4 2 - 2}{7} \cdot 27 = 2 42 - 2

\frac{4 2 - 2}{7} \cdot 27

使用根式运算法则:

a b = a \cdot b

7 \frac{4 2 - 2}{7} = 7 \cdot \frac{4 2 - 2}{7}

= 2 7 \cdot \frac{4 2 - 2}{7}

\frac{4 2 - 2}{7} \cdot 7 = 42 - 2

\frac{4 2 - 2}{7} \cdot 7

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 4 2 - 2 ) \cdot 7}{7}

约分:

7

= 42 - 2

= 2 42 - 2

= \frac{2 4 2 - 2 + 2 2 - 1}{2 7}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7}

有理化:

\frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

\frac{2 4 2 - 2 + 2 2 - 1}{2 7}

乘以共轭根式

\frac{7}{7}

= \frac{( 4 2 - 2 \cdot 2 + 2 2 - 1 ) 7}{2 7 7}

277 = 14

277

使用根式运算法则:

a a = a

77 = 7

= 2 \cdot 7

数字相乘:

2 \cdot 7 = 14

= 14

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14}

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

= \frac{7 ( 2 4 2 - 2 + 2 2 - 1 )}{14} - \frac{2 2 - 1}{2} i

无解

合并所有解

x \in R 无解

作图

流行的例子

sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x)

7 \cdot sin^{2} (x) + cos^{2} (x) - 1 = 6 sin (x)

4sec^2(x)-3tan^2(x)=5tan^2(x)

4 sec^{2} (x) - 3 tan^{2} (x) = 5 tan^{2} (x)

cos^{3} (x) = 66

2sqrt(3)*sin(4x+60^0)-3=0

23 \cdot sin (4 x + 6 0^{0}) - 3 = 0

(sin(x)+sin^2(x))/2 =0.5

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5