zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的代数 >

solvefor x,v(4-v(4+v(4-x)))=x^2-4

解答

求解

x, v (4 - v (4 + v (4 - x))) = x^{2} - 4

解答

x = \frac{v ^{3} + v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}, x = \frac{v ^{3} - v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}

求解步骤

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

展开

v (4 - v (4 + v (4 - x))) : 4 v - 4 v^{2} - 4 v^{3} + x v^{3}

4 v - 4 v^{2} - 4 v^{3} + x v^{3} = x^{2} - 4

交换两边

x^{2} - 4 = 4 v - 4 v^{2} - 4 v^{3} + x v^{3}

两边减去

- 4 v^{3} + x v^{3}

x^{2} - 4 - (- 4 v^{3} + x v^{3}) = 4 v - 4 v^{2} - 4 v^{3} + x v^{3} - (- 4 v^{3} + x v^{3})

化简

x^{2} - 4 + 4 v^{3} - x v^{3} = 4 v - 4 v^{2}

将

4 v^{2}

para o lado esquerdo

x^{2} - 4 + 4 v^{3} - x v^{3} + 4 v^{2} = 4 v

将

4 v

para o lado esquerdo

x^{2} - 4 + 4 v^{3} - x v^{3} + 4 v^{2} - 4 v = 0

改写成标准形式

a x^{2} + b x + c = 0

x^{2} - v^{3} x - 4 + 4 v^{3} + 4 v^{2} - 4 v = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - v ^{3} ) \pm ( - v ^{3} ) ^{2} - 4 \cdot 1 \cdot ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1}

化简

(- v^{3})^{2} - 4 \cdot 1 \cdot (- 4 + 4 v^{3} + 4 v^{2} - 4 v) : v^{6} - 4 (- 4 + 4 v^{3} + 4 v^{2} - 4 v)

x_{1, 2} = \frac{- ( - v ^{3} ) \pm v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - v ^{3} ) + v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1}, x_{2} = \frac{- ( - v ^{3} ) - v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1}

x = \frac{- ( - v ^{3} ) + v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1} : \frac{v ^{3} + v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}

x = \frac{- ( - v ^{3} ) - v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2 \cdot 1} : \frac{v ^{3} - v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}

二次方程组的解是:

x = \frac{v ^{3} + v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}, x = \frac{v ^{3} - v ^{6} - 4 ( - 4 + 4 v ^{3} + 4 v ^{2} - 4 v )}{2}

作图

流行的例子

d^3-d^2-6d=1+x^2

d^{3} - d^{2} - 6 d = 1 + x^{2}

3x^2-13x-10=(5-2)/3

3 x^{2} - 13 x - 10 = \frac{5 - 2}{3}

solvefor x,(x-1)^2+(y+3)^2+3=0

so l v e f or x, (x - 1)^{2} + (y + 3)^{2} + 3 = 0

solvefor y,x^2+y^2+xy=x

so l v e f or y, x^{2} + y^{2} + x y = x

solvefor x,x^2+y^2-xy=4

so l v e f or x, x^{2} + y^{2} - x y = 4