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受欢迎的三角函数 >

12cosh(2x)+7sinh(x)-24=0

解答

12 cosh (2 x) + 7 sinh (x) - 24 = 0

解答

x = ln (1.73025 \dots), x = ln (0.45623 \dots)

+1

度数

x = 31.41360 \dots^{\circ}, x = - 44.96326 \dots^{\circ}

求解步骤

12 cosh (2 x) + 7 sinh (x) - 24 = 0

使用三角恒等式改写

12 cosh (2 x) + 7 sinh (x) - 24 = 0

使用双曲函数恒等式:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

12 cosh (2 x) + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0

使用双曲函数恒等式:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

12 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0

12 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0

12 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0 : x = ln (1.73025 \dots), x = ln (0.45623 \dots)

12 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0

使用指数运算法则

12 \cdot \frac{e ^{2 x} + e ^{- 2 x}}{2} + 7 \cdot \frac{e ^{x} - e ^{- x}}{2} - 24 = 0

使用指数法则:

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

e^{2 x} = (e^{x})^{2}, e^{- 2 x} = (e^{x})^{- 2}, e^{- x} = (e^{x})^{- 1}

12 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} - 24 = 0

12 \cdot \frac{( e ^{x} ) ^{2} + ( e ^{x} ) ^{- 2}}{2} + 7 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} - 24 = 0

用

e^{x} = u

改写方程式

12 \cdot \frac{( u ) ^{2} + ( u ) ^{- 2}}{2} + 7 \cdot \frac{u - ( u ) ^{- 1}}{2} - 24 = 0

解

12 \cdot \frac{u ^{2} + u ^{- 2}}{2} + 7 \cdot \frac{u - u ^{- 1}}{2} - 24 = 0 : u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

12 \cdot \frac{u ^{2} + u ^{- 2}}{2} + 7 \cdot \frac{u - u ^{- 1}}{2} - 24 = 0

整理后得

\frac{6 ( u ^{4} + 1 )}{u ^{2}} + \frac{7 ( u ^{2} - 1 )}{2 u} - 24 = 0

乘以最小公倍数

\frac{6 ( u ^{4} + 1 )}{u ^{2}} + \frac{7 ( u ^{2} - 1 )}{2 u} - 24 = 0

找到

u^{2}, 2 u

的最小公倍数:

2 u^{2}

u^{2}, 2 u

最小公倍数 (LCM)

计算出由出现在

u^{2}

或

2 u

中的因子组成的表达式

= 2 u^{2}

乘以最小公倍数=

2 u^{2}

\frac{6 ( u ^{4} + 1 )}{u ^{2}} \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} - 24 \cdot 2 u^{2} = 0 \cdot 2 u^{2}

化简

\frac{6 ( u ^{4} + 1 )}{u ^{2}} \cdot 2 u^{2} + \frac{7 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} - 24 \cdot 2 u^{2} = 0 \cdot 2 u^{2}

化简

\frac{6 ( u ^{4} + 1 )}{u ^{2}} \cdot 2 u^{2} : 12 (u^{4} + 1)

\frac{6 ( u ^{4} + 1 )}{u ^{2}} \cdot 2 u^{2}

分式相乘:

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

= \frac{6 ( u ^{4} + 1 ) \cdot 2 u ^{2}}{u ^{2}}

约分:

u^{2}

= 6 (u^{4} + 1) \cdot 2

数字相乘:

6 \cdot 2 = 12

= 12 (u^{4} + 1)

化简

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

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

分式相乘:

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

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

约分:

2

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

约分:

u

= 7 u (u^{2} - 1)

化简

- 24 \cdot 2 u^{2} : - 48 u^{2}

- 24 \cdot 2 u^{2}

数字相乘:

24 \cdot 2 = 48

= - 48 u^{2}

化简

0 \cdot 2 u^{2} : 0

0 \cdot 2 u^{2}

使用法则

0 \cdot a = 0

= 0

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} = 0

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} = 0

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} = 0

解

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} = 0 : u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} = 0

展开

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2} : 12 u^{4} + 12 + 7 u^{3} - 7 u - 48 u^{2}

12 (u^{4} + 1) + 7 u (u^{2} - 1) - 48 u^{2}

乘开

12 (u^{4} + 1) : 12 u^{4} + 12

12 (u^{4} + 1)

使用分配律:

a (b + c) = ab + a c

a = 12, b = u^{4}, c = 1

= 12 u^{4} + 12 \cdot 1

数字相乘:

12 \cdot 1 = 12

= 12 u^{4} + 12

= 12 u^{4} + 12 + 7 u (u^{2} - 1) - 48 u^{2}

乘开

7 u (u^{2} - 1) : 7 u^{3} - 7 u

7 u (u^{2} - 1)

使用分配律:

a (b - c) = ab - a c

a = 7 u, b = u^{2}, c = 1

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

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

化简

7 u^{2} u - 7 \cdot 1 \cdot u : 7 u^{3} - 7 u

7 u^{2} u - 7 \cdot 1 \cdot u

7 u^{2} u = 7 u^{3}

7 u^{2} u

使用指数法则:

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

u^{2} u = u^{2 + 1}

= 7 u^{2 + 1}

数字相加:

2 + 1 = 3

= 7 u^{3}

7 \cdot 1 \cdot u = 7 u

7 \cdot 1 \cdot u

数字相乘:

7 \cdot 1 = 7

= 7 u

= 7 u^{3} - 7 u

= 7 u^{3} - 7 u

= 12 u^{4} + 12 + 7 u^{3} - 7 u - 48 u^{2}

12 u^{4} + 12 + 7 u^{3} - 7 u - 48 u^{2} = 0

改写成标准形式

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

12 u^{4} + 7 u^{3} - 48 u^{2} - 7 u + 12 = 0

使用牛顿-拉弗森方法找到

12 u^{4} + 7 u^{3} - 48 u^{2} - 7 u + 12 = 0

的一个解:

u \approx 1.73025 \dots

12 u^{4} + 7 u^{3} - 48 u^{2} - 7 u + 12 = 0

牛顿-拉弗森近似法定义

f (u) = 12 u^{4} + 7 u^{3} - 48 u^{2} - 7 u + 12

找到

f^{^{'}} (u) : 48 u^{3} + 21 u^{2} - 96 u - 7

\frac{d}{d u} (12 u^{4} + 7 u^{3} - 48 u^{2} - 7 u + 12)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (12 u^{4}) + \frac{d}{d u} (7 u^{3}) - \frac{d}{d u} (48 u^{2}) - \frac{d}{d u} (7 u) + \frac{d}{d u} (12)

\frac{d}{d u} (12 u^{4}) = 48 u^{3}

\frac{d}{d u} (12 u^{4})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 12 \frac{d}{d u} (u^{4})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 12 \cdot 4 u^{4 - 1}

化简

= 48 u^{3}

\frac{d}{d u} (7 u^{3}) = 21 u^{2}

\frac{d}{d u} (7 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 7 \cdot 3 u^{3 - 1}

化简

= 21 u^{2}

\frac{d}{d u} (48 u^{2}) = 96 u

\frac{d}{d u} (48 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 48 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 48 \cdot 2 u^{2 - 1}

化简

= 96 u

\frac{d}{d u} (7 u) = 7

\frac{d}{d u} (7 u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 7 \cdot 1

化简

= 7

\frac{d}{d u} (12) = 0

\frac{d}{d u} (12)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 48 u^{3} + 21 u^{2} - 96 u - 7 + 0

化简

= 48 u^{3} + 21 u^{2} - 96 u - 7

令

u_{0} = 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 1.79925 \dots : Δ u_{1} = 0.20074 \dots

f (u_{0}) = 12 \cdot 2^{4} + 7 \cdot 2^{3} - 48 \cdot 2^{2} - 7 \cdot 2 + 12 = 54

f^{^{'}} (u_{0}) = 48 \cdot 2^{3} + 21 \cdot 2^{2} - 96 \cdot 2 - 7 = 269

u_{1} = 1.79925 \dots

Δ u_{1} = ∣ 1.79925 \dots - 2 ∣ = 0.20074 \dots

Δ u_{1} = 0.20074 \dots

u_{2} = 1.73639 \dots : Δ u_{2} = 0.06285 \dots

f (u_{1}) = 12 \cdot 1.79925 \dots^{4} + 7 \cdot 1.79925 \dots^{3} - 48 \cdot 1.79925 \dots^{2} - 7 \cdot 1.79925 \dots + 12 = 10.55028 \dots

f^{^{'}} (u_{1}) = 48 \cdot 1.79925 \dots^{3} + 21 \cdot 1.79925 \dots^{2} - 96 \cdot 1.79925 \dots - 7 = 167.84443 \dots

u_{2} = 1.73639 \dots

Δ u_{2} = ∣ 1.73639 \dots - 1.79925 \dots ∣ = 0.06285 \dots

Δ u_{2} = 0.06285 \dots

u_{3} = 1.73031 \dots : Δ u_{3} = 0.00608 \dots

f (u_{2}) = 12 \cdot 1.73639 \dots^{4} + 7 \cdot 1.73639 \dots^{3} - 48 \cdot 1.73639 \dots^{2} - 7 \cdot 1.73639 \dots + 12 = 0.85758 \dots

f^{^{'}} (u_{2}) = 48 \cdot 1.73639 \dots^{3} + 21 \cdot 1.73639 \dots^{2} - 96 \cdot 1.73639 \dots - 7 = 140.92085 \dots

u_{3} = 1.73031 \dots

Δ u_{3} = ∣ 1.73031 \dots - 1.73639 \dots ∣ = 0.00608 \dots

Δ u_{3} = 0.00608 \dots

u_{4} = 1.73025 \dots : Δ u_{4} = 0.00005 \dots

f (u_{3}) = 12 \cdot 1.73031 \dots^{4} + 7 \cdot 1.73031 \dots^{3} - 48 \cdot 1.73031 \dots^{2} - 7 \cdot 1.73031 \dots + 12 = 0.00759 \dots

f^{^{'}} (u_{3}) = 48 \cdot 1.73031 \dots^{3} + 21 \cdot 1.73031 \dots^{2} - 96 \cdot 1.73031 \dots - 7 = 138.42910 \dots

u_{4} = 1.73025 \dots

Δ u_{4} = ∣ 1.73025 \dots - 1.73031 \dots ∣ = 0.00005 \dots

Δ u_{4} = 0.00005 \dots

u_{5} = 1.73025 \dots : Δ u_{5} = 4.43112 E - 9

f (u_{4}) = 12 \cdot 1.73025 \dots^{4} + 7 \cdot 1.73025 \dots^{3} - 48 \cdot 1.73025 \dots^{2} - 7 \cdot 1.73025 \dots + 12 = 6.13297 E - 7

f^{^{'}} (u_{4}) = 48 \cdot 1.73025 \dots^{3} + 21 \cdot 1.73025 \dots^{2} - 96 \cdot 1.73025 \dots - 7 = 138.40674 \dots

u_{5} = 1.73025 \dots

Δ u_{5} = ∣ 1.73025 \dots - 1.73025 \dots ∣ = 4.43112 E - 9

Δ u_{5} = 4.43112 E - 9

u \approx 1.73025 \dots

使用长除法

Eq u a t i o n 0 : \frac{12 u ^{4} + 7 u ^{3} - 48 u ^{2} - 7 u + 12}{u - 1.73025 \dots} = 12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots

12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots \approx 0

使用牛顿-拉弗森方法找到

12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots = 0

的一个解:

u \approx 0.45623 \dots

12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots

找到

f^{^{'}} (u) : 36 u^{2} + 55.52620 \dots u + 0.03734 \dots

\frac{d}{d u} (12 u^{3} + 27.76310 \dots u^{2} + 0.03734 \dots u - 6.93537 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (12 u^{3}) + \frac{d}{d u} (27.76310 \dots u^{2}) + \frac{d}{d u} (0.03734 \dots u) - \frac{d}{d u} (6.93537 \dots)

\frac{d}{d u} (12 u^{3}) = 36 u^{2}

\frac{d}{d u} (12 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 12 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 12 \cdot 3 u^{3 - 1}

化简

= 36 u^{2}

\frac{d}{d u} (27.76310 \dots u^{2}) = 55.52620 \dots u

\frac{d}{d u} (27.76310 \dots u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 27.76310 \dots \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 27.76310 \dots \cdot 2 u^{2 - 1}

化简

= 55.52620 \dots u

\frac{d}{d u} (0.03734 \dots u) = 0.03734 \dots

\frac{d}{d u} (0.03734 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.03734 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 0.03734 \dots \cdot 1

化简

= 0.03734 \dots

\frac{d}{d u} (6.93537 \dots) = 0

\frac{d}{d u} (6.93537 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 36 u^{2} + 55.52620 \dots u + 0.03734 \dots - 0

化简

= 36 u^{2} + 55.52620 \dots u + 0.03734 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.6410682029 : Δ u_{1} = 0.3589317971

f (u_{0}) = 12 \cdot 1^{3} + 27.76310 \dots \cdot 1^{2} + 0.03734 \dots \cdot 1 - 6.93537 \dots = 32.86507 \dots

f^{^{'}} (u_{0}) = 36 \cdot 1^{2} + 55.52620 \dots \cdot 1 + 0.03734 \dots = 91.56355 \dots

u_{1} = 0.6410682029

Δ u_{1} = ∣ 0.6410682029 - 1 ∣ = 0.3589317971

Δ u_{1} = 0.3589317971

u_{2} = 0.48917 \dots : Δ u_{2} = 0.15189 \dots

f (u_{1}) = 12 \cdot 0.641068202 9^{3} + 27.76310 \dots \cdot 0.641068202 9^{2} + 0.03734 \dots \cdot 0.6410682029 - 6.93537 \dots = 7.65982 \dots

f^{^{'}} (u_{1}) = 36 \cdot 0.641068202 9^{2} + 55.52620 \dots \cdot 0.6410682029 + 0.03734 \dots = 50.42829 \dots

u_{2} = 0.48917 \dots

Δ u_{2} = ∣ 0.48917 \dots - 0.6410682029 ∣ = 0.15189 \dots

Δ u_{2} = 0.15189 \dots

u_{3} = 0.45759 \dots : Δ u_{3} = 0.03157 \dots

f (u_{2}) = 12 \cdot 0.48917 \dots^{3} + 27.76310 \dots \cdot 0.48917 \dots^{2} + 0.03734 \dots \cdot 0.48917 \dots - 6.93537 \dots = 1.13097 \dots

f^{^{'}} (u_{2}) = 36 \cdot 0.48917 \dots^{2} + 55.52620 \dots \cdot 0.48917 \dots + 0.03734 \dots = 35.81369 \dots

u_{3} = 0.45759 \dots

Δ u_{3} = ∣ 0.45759 \dots - 0.48917 \dots ∣ = 0.03157 \dots

Δ u_{3} = 0.03157 \dots

u_{4} = 0.45623 \dots : Δ u_{4} = 0.00136 \dots

f (u_{3}) = 12 \cdot 0.45759 \dots^{3} + 27.76310 \dots \cdot 0.45759 \dots^{2} + 0.03734 \dots \cdot 0.45759 \dots - 6.93537 \dots = 0.04487 \dots

f^{^{'}} (u_{3}) = 36 \cdot 0.45759 \dots^{2} + 55.52620 \dots \cdot 0.45759 \dots + 0.03734 \dots = 32.98387 \dots

u_{4} = 0.45623 \dots

Δ u_{4} = ∣ 0.45623 \dots - 0.45759 \dots ∣ = 0.00136 \dots

Δ u_{4} = 0.00136 \dots

u_{5} = 0.45623 \dots : Δ u_{5} = 2.49018 E - 6

f (u_{4}) = 12 \cdot 0.45623 \dots^{3} + 27.76310 \dots \cdot 0.45623 \dots^{2} + 0.03734 \dots \cdot 0.45623 \dots - 6.93537 \dots = 0.00008 \dots

f^{^{'}} (u_{4}) = 36 \cdot 0.45623 \dots^{2} + 55.52620 \dots \cdot 0.45623 \dots + 0.03734 \dots = 32.86358 \dots

u_{5} = 0.45623 \dots

Δ u_{5} = ∣ 0.45623 \dots - 0.45623 \dots ∣ = 2.49018 E - 6

Δ u_{5} = 2.49018 E - 6

u_{6} = 0.45623 \dots : Δ u_{6} = 8.33775 E - 12

f (u_{5}) = 12 \cdot 0.45623 \dots^{3} + 27.76310 \dots \cdot 0.45623 \dots^{2} + 0.03734 \dots \cdot 0.45623 \dots - 6.93537 \dots = 2.74007 E - 10

f^{^{'}} (u_{5}) = 36 \cdot 0.45623 \dots^{2} + 55.52620 \dots \cdot 0.45623 \dots + 0.03734 \dots = 32.86336 \dots

u_{6} = 0.45623 \dots

Δ u_{6} = ∣ 0.45623 \dots - 0.45623 \dots ∣ = 8.33775 E - 12

Δ u_{6} = 8.33775 E - 12

u \approx 0.45623 \dots

使用长除法

Eq u a t i o n 0 : \frac{12 u ^{3} + 27.76310 \dots u ^{2} + 0.03734 \dots u - 6.93537 \dots}{u - 0.45623 \dots} = 12 u^{2} + 33.23786 \dots u + 15.20147 \dots

12 u^{2} + 33.23786 \dots u + 15.20147 \dots \approx 0

使用牛顿-拉弗森方法找到

12 u^{2} + 33.23786 \dots u + 15.20147 \dots = 0

的一个解:

u \approx - 0.57794 \dots

12 u^{2} + 33.23786 \dots u + 15.20147 \dots = 0

牛顿-拉弗森近似法定义

f (u) = 12 u^{2} + 33.23786 \dots u + 15.20147 \dots

找到

f^{^{'}} (u) : 24 u + 33.23786 \dots

\frac{d}{d u} (12 u^{2} + 33.23786 \dots u + 15.20147 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (12 u^{2}) + \frac{d}{d u} (33.23786 \dots u) + \frac{d}{d u} (15.20147 \dots)

\frac{d}{d u} (12 u^{2}) = 24 u

\frac{d}{d u} (12 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 12 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 12 \cdot 2 u^{2 - 1}

化简

= 24 u

\frac{d}{d u} (33.23786 \dots u) = 33.23786 \dots

\frac{d}{d u} (33.23786 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 33.23786 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 33.23786 \dots \cdot 1

化简

= 33.23786 \dots

\frac{d}{d u} (15.20147 \dots) = 0

\frac{d}{d u} (15.20147 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 24 u + 33.23786 \dots + 0

化简

= 24 u + 33.23786 \dots

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.45735 \dots : Δ u_{1} = 0.45735 \dots

f (u_{0}) = 12 \cdot 0^{2} + 33.23786 \dots \cdot 0 + 15.20147 \dots = 15.20147 \dots

f^{^{'}} (u_{0}) = 24 \cdot 0 + 33.23786 \dots = 33.23786 \dots

u_{1} = - 0.45735 \dots

Δ u_{1} = ∣ - 0.45735 \dots - 0 ∣ = 0.45735 \dots

Δ u_{1} = 0.45735 \dots

u_{2} = - 0.57010 \dots : Δ u_{2} = 0.11275 \dots

f (u_{1}) = 12 (- 0.45735 \dots)^{2} + 33.23786 \dots (- 0.45735 \dots) + 15.20147 \dots = 2.51007 \dots

f^{^{'}} (u_{1}) = 24 (- 0.45735 \dots) + 33.23786 \dots = 22.26136 \dots

u_{2} = - 0.57010 \dots

Δ u_{2} = ∣ - 0.57010 \dots - (- 0.45735 \dots) ∣ = 0.11275 \dots

Δ u_{2} = 0.11275 \dots

u_{3} = - 0.57791 \dots : Δ u_{3} = 0.00780 \dots

f (u_{2}) = 12 (- 0.57010 \dots)^{2} + 33.23786 \dots (- 0.57010 \dots) + 15.20147 \dots = 0.15256 \dots

f^{^{'}} (u_{2}) = 24 (- 0.57010 \dots) + 33.23786 \dots = 19.55525 \dots

u_{3} = - 0.57791 \dots

Δ u_{3} = ∣ - 0.57791 \dots - (- 0.57010 \dots) ∣ = 0.00780 \dots

Δ u_{3} = 0.00780 \dots

u_{4} = - 0.57794 \dots : Δ u_{4} = 0.00003 \dots

f (u_{3}) = 12 (- 0.57791 \dots)^{2} + 33.23786 \dots (- 0.57791 \dots) + 15.20147 \dots = 0.00073 \dots

f^{^{'}} (u_{3}) = 24 (- 0.57791 \dots) + 33.23786 \dots = 19.36801 \dots

u_{4} = - 0.57794 \dots

Δ u_{4} = ∣ - 0.57794 \dots - (- 0.57791 \dots) ∣ = 0.00003 \dots

Δ u_{4} = 0.00003 \dots

u_{5} = - 0.57794 \dots : Δ u_{5} = 8.81165 E - 10

f (u_{4}) = 12 (- 0.57794 \dots)^{2} + 33.23786 \dots (- 0.57794 \dots) + 15.20147 \dots = 1.70656 E - 8

f^{^{'}} (u_{4}) = 24 (- 0.57794 \dots) + 33.23786 \dots = 19.36711 \dots

u_{5} = - 0.57794 \dots

Δ u_{5} = ∣ - 0.57794 \dots - (- 0.57794 \dots) ∣ = 8.81165 E - 10

Δ u_{5} = 8.81165 E - 10

u \approx - 0.57794 \dots

使用长除法

Eq u a t i o n 0 : \frac{12 u ^{2} + 33.23786 \dots u + 15.20147 \dots}{u + 0.57794 \dots} = 12 u + 26.30249 \dots

12 u + 26.30249 \dots \approx 0

u \approx - 2.19187 \dots

解为

u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

12 \frac{u ^{2} + u ^{- 2}}{2} + 7 \frac{u - u ^{- 1}}{2} - 24

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

u \approx 1.73025 \dots, u \approx 0.45623 \dots, u \approx - 0.57794 \dots, u \approx - 2.19187 \dots

代回

u = e^{x}

，求解

x

解

e^{x} = 1.73025 \dots : x = ln (1.73025 \dots)

e^{x} = 1.73025 \dots

使用指数运算法则

e^{x} = 1.73025 \dots

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1.73025 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1.73025 \dots)

x = ln (1.73025 \dots)

解

e^{x} = 0.45623 \dots : x = ln (0.45623 \dots)

e^{x} = 0.45623 \dots

使用指数运算法则

e^{x} = 0.45623 \dots

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (0.45623 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.45623 \dots)

x = ln (0.45623 \dots)

解

e^{x} = - 0.57794 \dots : x \in R

无解

e^{x} = - 0.57794 \dots

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

e^{x} = - 2.19187 \dots : x \in R

无解

e^{x} = - 2.19187 \dots

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = ln (1.73025 \dots), x = ln (0.45623 \dots)

x = ln (1.73025 \dots), x = ln (0.45623 \dots)

作图

流行的例子

sec^{2} (θ) - 1 = 0

2sin(θ)cos(θ)=-sin(θ)

2 sin (θ) cos (θ) = - sin (θ)

sec (θ) = \frac{7}{6}

2cos^2(x)+7sin(x)-5=0

2 cos^{2} (x) + 7 sin (x) - 5 = 0

8tan(2x)-16cos(x)=0

8 tan (2 x) - 16 cos (x) = 0