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受欢迎的代数 >

1/4 (2A)+2Ax+B+Ax^2+Bx+C=x^2-2x

解答

\frac{1}{4} (2 A) + 2 A x + B + A x^{2} + B x + C = x^{2} - 2 x

解答

x = - \frac{2 A + B + 2 - 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4}{2 ( A - 1 )}, x = - \frac{2 A + B + 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4 + 2}{2 ( A - 1 )}; A \neq = 1

求解步骤

\frac{1}{4} (2 A) + 2 A x + B + A x^{2} + B x + C = x^{2} - 2 x

在两边乘以

4

2 A + 8 A x + 4 B + 4 A x^{2} + 4 B x + 4 C = 4 x^{2} - 8 x

将

8 x

para o lado esquerdo

2 A + 8 A x + 4 B + 4 A x^{2} + 4 B x + 4 C + 8 x = 4 x^{2}

将

4 x^{2}

para o lado esquerdo

2 A + 8 A x + 4 B + 4 A x^{2} + 4 B x + 4 C + 8 x - 4 x^{2} = 0

改写成标准形式

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

(4 A - 4) x^{2} + (8 A + 4 B + 8) x + 2 A + 4 B + 4 C = 0

使用求根公式求解

x_{1, 2} = \frac{- ( 8 A + 4 B + 8 ) \pm ( 8 A + 4 B + 8 ) ^{2} - 4 ( 4 A - 4 ) ( 2 A + 4 B + 4 C )}{2 ( 4 A - 4 )}

化简

(8 A + 4 B + 8)^{2} - 4 (4 A - 4) (2 A + 4 B + 4 C) : 4 2 A^{2} - 4 A C + 10 A + B^{2} + 8 B + 4 C + 4

x_{1, 2} = \frac{- ( 8 A + 4 B + 8 ) \pm 4 2 A ^{2} - 4 A C + 10 A + B ^{2} + 8 B + 4 C + 4}{2 ( 4 A - 4 )}; A \neq = 1

将解分隔开

x_{1} = \frac{- ( 8 A + 4 B + 8 ) + 4 2 A ^{2} - 4 A C + 10 A + B ^{2} + 8 B + 4 C + 4}{2 ( 4 A - 4 )}, x_{2} = \frac{- ( 8 A + 4 B + 8 ) - 4 2 A ^{2} - 4 A C + 10 A + B ^{2} + 8 B + 4 C + 4}{2 ( 4 A - 4 )}

x = \frac{- ( 8 A + 4 B + 8 ) + 4 2 A ^{2} - 4 A C + 10 A + B ^{2} + 8 B + 4 C + 4}{2 ( 4 A - 4 )} : - \frac{2 A + B + 2 - 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4}{2 ( A - 1 )}

x = \frac{- ( 8 A + 4 B + 8 ) - 4 2 A ^{2} - 4 A C + 10 A + B ^{2} + 8 B + 4 C + 4}{2 ( 4 A - 4 )} : - \frac{2 A + B + 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4 + 2}{2 ( A - 1 )}

二次方程组的解是:

x = - \frac{2 A + B + 2 - 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4}{2 ( A - 1 )}, x = - \frac{2 A + B + 2 A ^{2} + 10 A - 4 A C + B ^{2} + 8 B + 4 C + 4 + 2}{2 ( A - 1 )}; A \neq = 1

作图

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