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受欢迎的代数 >

a(x-11)(x+a)=x^2+0x-11a

解答

a (x - 11) (x + a) = x^{2} + 0 x - 11 a

解答

x = \frac{- a ^{2} + 11 a + a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}, x = \frac{- a ^{2} + 11 a - a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}; a \neq = 1

求解步骤

a (x - 11) (x + a) = x^{2} + 0 \cdot x - 11 a

展开

a (x - 11) (x + a) : a x^{2} + a^{2} x - 11 a x - 11 a^{2}

展开

x^{2} + 0 \cdot x - 11 a : x^{2} - 11 a

a x^{2} + a^{2} x - 11 a x - 11 a^{2} = x^{2} - 11 a

将

11 a

para o lado esquerdo

a x^{2} + a^{2} x - 11 a x - 11 a^{2} + 11 a = x^{2}

将

x^{2}

para o lado esquerdo

a x^{2} + a^{2} x - 11 a x - 11 a^{2} + 11 a - x^{2} = 0

改写成标准形式

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

(a - 1) x^{2} + (a^{2} - 11 a) x - 11 a^{2} + 11 a = 0

使用求根公式求解

x_{1, 2} = \frac{- ( a ^{2} - 11 a ) \pm ( a ^{2} - 11 a ) ^{2} - 4 ( a - 1 ) ( - 11 a ^{2} + 11 a )}{2 ( a - 1 )}

化简

(a^{2} - 11 a)^{2} - 4 (a - 1) (- 11 a^{2} + 11 a) : a^{4} + 22 a^{3} + 33 a^{2} + 44 a

x_{1, 2} = \frac{- ( a ^{2} - 11 a ) \pm a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}; a \neq = 1

将解分隔开

x_{1} = \frac{- ( a ^{2} - 11 a ) + a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}, x_{2} = \frac{- ( a ^{2} - 11 a ) - a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}

x = \frac{- ( a ^{2} - 11 a ) + a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )} : \frac{- a ^{2} + 11 a + a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}

x = \frac{- ( a ^{2} - 11 a ) - a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )} : \frac{- a ^{2} + 11 a - a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}

二次方程组的解是:

x = \frac{- a ^{2} + 11 a + a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}, x = \frac{- a ^{2} + 11 a - a ^{4} + 22 a ^{3} + 33 a ^{2} + 44 a}{2 ( a - 1 )}; a \neq = 1

作图

流行的例子

5 x^{2} + 7 x - 3 = - x^{2}

(x + 5)^{2} + 12 = 76

2 x^{2} + 6 + 4 = 0

solvefor x,z^2=y^2+x^2

so l v e f or x, z^{2} = y^{2} + x^{2}

12 x^{2} - 5 = - 3