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,f(x+4)=x^2-2x+7

解答

求解

x, f (x + 4) = x^{2} - 2 x + 7

解答

x = \frac{2 + f + f ^{2} + 20 f - 24}{2}, x = \frac{2 + f - f ^{2} + 20 f - 24}{2}

求解步骤

f (x + 4) = x^{2} - 2 x + 7

展开

f (x + 4) : x f + 4 f

x f + 4 f = x^{2} - 2 x + 7

交换两边

x^{2} - 2 x + 7 = x f + 4 f

将

4 f

para o lado esquerdo

x^{2} - 2 x + 7 - 4 f = x f

将

x f

para o lado esquerdo

x^{2} - 2 x + 7 - 4 f - x f = 0

改写成标准形式

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

x^{2} - (2 + f) x + 7 - 4 f = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 2 - f ) \pm ( - 2 - f ) ^{2} - 4 \cdot 1 \cdot ( 7 - 4 f )}{2 \cdot 1}

化简

(- 2 - f)^{2} - 4 \cdot 1 \cdot (7 - 4 f) : f^{2} + 20 f - 24

x_{1, 2} = \frac{- ( - 2 - f ) \pm f ^{2} + 20 f - 24}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 2 - f ) + f ^{2} + 20 f - 24}{2 \cdot 1}, x_{2} = \frac{- ( - 2 - f ) - f ^{2} + 20 f - 24}{2 \cdot 1}

x = \frac{- ( - 2 - f ) + f ^{2} + 20 f - 24}{2 \cdot 1} : \frac{2 + f + f ^{2} + 20 f - 24}{2}

x = \frac{- ( - 2 - f ) - f ^{2} + 20 f - 24}{2 \cdot 1} : \frac{2 + f - f ^{2} + 20 f - 24}{2}

二次方程组的解是:

x = \frac{2 + f + f ^{2} + 20 f - 24}{2}, x = \frac{2 + f - f ^{2} + 20 f - 24}{2}

作图

流行的例子

6 x (x - 1) = 40 - 5 x

x^2+(m-3)x+m+1=0

x^{2} + (m - 3) x + m + 1 = 0

-5b^2+9b+13=-6b^2

- 5 b^{2} + 9 b + 13 = - 6 b^{2}

- x^{2} + 5 x + 4 = 6

- 8 + 13 x^{2} = 0