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 >

(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0

解答

(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0

解答

x = \frac{b + c + a + b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}, x = \frac{b + c + a - b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}

求解步骤

(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0

展开

(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) : 3 x^{2} - 2 b x - 2 c x - 2 a x + b c + c a + ba

3 x^{2} - 2 b x - 2 c x - 2 a x + b c + c a + ba = 0

改写成标准形式

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

3 x^{2} - (2 b + 2 c + 2 a) x + b c + c a + ba = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 2 b - 2 c - 2 a ) \pm ( - 2 b - 2 c - 2 a ) ^{2} - 4 \cdot 3 ( b c + c a + ba )}{2 \cdot 3}

化简

(- 2 b - 2 c - 2 a)^{2} - 4 \cdot 3 (b c + c a + ba) : 2 b^{2} - b c - ba + c^{2} + a^{2} - c a

x_{1, 2} = \frac{- ( - 2 b - 2 c - 2 a ) \pm 2 b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{2 \cdot 3}

将解分隔开

x_{1} = \frac{- ( - 2 b - 2 c - 2 a ) + 2 b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{2 \cdot 3}, x_{2} = \frac{- ( - 2 b - 2 c - 2 a ) - 2 b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{2 \cdot 3}

x = \frac{- ( - 2 b - 2 c - 2 a ) + 2 b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{2 \cdot 3} : \frac{b + c + a + b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}

x = \frac{- ( - 2 b - 2 c - 2 a ) - 2 b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{2 \cdot 3} : \frac{b + c + a - b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}

二次方程组的解是:

x = \frac{b + c + a + b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}, x = \frac{b + c + a - b ^{2} - b c - ba + c ^{2} + a ^{2} - c a}{3}

作图

流行的例子

solvefor x,15(2x^2-y^2)=7xy

so l v e f or x, 15 (2 x^{2} - y^{2}) = 7 x y

(3k+1)x^2+(2k+2)x+k=0

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

(6 x + 2)^{2} = 16

((x+2)^2-(2x+11))/3 =3

\frac{( x + 2 ) ^{2} - ( 2 x + 11 )}{3} = 3

3/1+x-x^2+1/1-x^2+x+1/1-x=0

\frac{3}{1} + x - x^{2} + \frac{1}{1} - x^{2} + x + \frac{1}{1} - x = 0