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受欢迎的代数 >

1=a(2x^2-x)+b(3x^2+1)+c(x^2)

解答

1 = a (2 x^{2} - x) + b (3 x^{2} + 1) + c (x^{2})

解答

x = \frac{a + a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}, x = \frac{a - a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}; 2 a + 3 b + c \neq = 0

求解步骤

1 = a (2 x^{2} - x) + b (3 x^{2} + 1) + c (x^{2})

展开

a (2 x^{2} - x) + b (3 x^{2} + 1) + c (x^{2}) : 2 a x^{2} - a x + 3 b x^{2} + b + c x^{2}

1 = 2 a x^{2} - a x + 3 b x^{2} + b + c x^{2}

交换两边

2 a x^{2} - a x + 3 b x^{2} + b + c x^{2} = 1

将

1

para o lado esquerdo

2 a x^{2} - a x + 3 b x^{2} + b + c x^{2} - 1 = 0

改写成标准形式

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

(2 a + 3 b + c) x^{2} - a x + b - 1 = 0

使用求根公式求解

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

化简

(- a)^{2} - 4 (2 a + 3 b + c) (b - 1) : a^{2} - 4 (2 a + 3 b + c) (b - 1)

x_{1, 2} = \frac{- ( - a ) \pm a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}; 2 a + 3 b + c \neq = 0

将解分隔开

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

x = \frac{- ( - a ) + a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )} : \frac{a + a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}

x = \frac{- ( - a ) - a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )} : \frac{a - a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}

二次方程组的解是:

x = \frac{a + a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}, x = \frac{a - a ^{2} - 4 ( 2 a + 3 b + c ) ( b - 1 )}{2 ( 2 a + 3 b + c )}; 2 a + 3 b + c \neq = 0

作图

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