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受欢迎的代数 >

(x^2+ax-10)(1-a)=(2a+6)(1-x)

解答

(x^{2} + a x - 10) (1 - a) = (2 a + 6) (1 - x)

解答

x = \frac{a ^{2} - 3 a - 6 + a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}, x = \frac{a ^{2} - 3 a - 6 - a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}; a \neq = 1

求解步骤

(x^{2} + a x - 10) (1 - a) = (2 a + 6) (1 - x)

展开

(x^{2} + a x - 10) (1 - a) : x^{2} - a x^{2} + a x - a^{2} x - 10 + 10 a

展开

(2 a + 6) (1 - x) : 2 a - 2 a x + 6 - 6 x

x^{2} - a x^{2} + a x - a^{2} x - 10 + 10 a = 2 a - 2 a x + 6 - 6 x

将

6 x

para o lado esquerdo

x^{2} - a x^{2} + a x - a^{2} x - 10 + 10 a + 6 x = 2 a - 2 a x + 6

将

6

para o lado esquerdo

x^{2} - a x^{2} + a x + 6 x - a^{2} x + 10 a - 16 = 2 a - 2 a x

两边减去

2 a - 2 a x

x^{2} - a x^{2} + a x + 6 x - a^{2} x + 10 a - 16 - (2 a - 2 a x) = 2 a - 2 a x - (2 a - 2 a x)

化简

(- a + 1) x^{2} + (- a^{2} + 3 a + 6) x + 8 a - 16 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - a ^{2} + 3 a + 6 ) \pm ( - a ^{2} + 3 a + 6 ) ^{2} - 4 ( - a + 1 ) ( 8 a - 16 )}{2 ( - a + 1 )}

化简

(- a^{2} + 3 a + 6)^{2} - 4 (- a + 1) (8 a - 16) : a^{4} - 6 a^{3} + 29 a^{2} - 60 a + 100

x_{1, 2} = \frac{- ( - a ^{2} + 3 a + 6 ) \pm a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}; a \neq = 1

将解分隔开

x_{1} = \frac{- ( - a ^{2} + 3 a + 6 ) + a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}, x_{2} = \frac{- ( - a ^{2} + 3 a + 6 ) - a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}

x = \frac{- ( - a ^{2} + 3 a + 6 ) + a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )} : \frac{a ^{2} - 3 a - 6 + a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}

x = \frac{- ( - a ^{2} + 3 a + 6 ) - a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )} : \frac{a ^{2} - 3 a - 6 - a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}

二次方程组的解是:

x = \frac{a ^{2} - 3 a - 6 + a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}, x = \frac{a ^{2} - 3 a - 6 - a ^{4} - 6 a ^{3} + 29 a ^{2} - 60 a + 100}{2 ( - a + 1 )}; a \neq = 1

作图

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