(a+b)^2x^2+8(a^2-b^2)x+16(a-b)^2=0

解答

(a + b)^{2} x^{2} + 8 (a^{2} - b^{2}) x + 16 (a - b)^{2} = 0

x = - \frac{4 ( a - b )}{a + b}; a^{2} + 2 ab + b^{2} \neq = 0

求解步骤

(a + b)^{2} x^{2} + 8 (a^{2} - b^{2}) x + 16 (a - b)^{2} = 0

展开

(a + b)^{2} x^{2} + 8 (a^{2} - b^{2}) x + 16 (a - b)^{2} : a^{2} x^{2} + 2 ab x^{2} + b^{2} x^{2} + 8 a^{2} x - 8 b^{2} x + 16 a^{2} - 32 ab + 16 b^{2}

a^{2} x^{2} + 2 ab x^{2} + b^{2} x^{2} + 8 a^{2} x - 8 b^{2} x + 16 a^{2} - 32 ab + 16 b^{2} = 0

改写成标准形式

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

(a^{2} + 2 ab + b^{2}) x^{2} + (8 a^{2} - 8 b^{2}) x + 16 a^{2} - 32 ab + 16 b^{2} = 0

使用求根公式求解

x_{1, 2} = \frac{- ( 8 a ^{2} - 8 b ^{2} ) \pm ( 8 a ^{2} - 8 b ^{2} ) ^{2} - 4 ( a ^{2} + 2 ab + b ^{2} ) ( 16 a ^{2} - 32 ab + 16 b ^{2} )}{2 ( a ^{2} + 2 ab + b ^{2} )}

(8 a^{2} - 8 b^{2})^{2} - 4 (a^{2} + 2 ab + b^{2}) (16 a^{2} - 32 ab + 16 b^{2}) = 0

x_{1, 2} = \frac{- ( 8 a ^{2} - 8 b ^{2} ) \pm 0}{2 ( a ^{2} + 2 ab + b ^{2} )}; a^{2} + 2 ab + b^{2} \neq = 0

x = \frac{- ( 8 a ^{2} - 8 b ^{2} )}{2 ( a ^{2} + 2 ab + b ^{2} )}; a^{2} + 2 ab + b^{2} \neq = 0

化简

\frac{- ( 8 a ^{2} - 8 b ^{2} )}{2 ( a ^{2} + 2 ab + b ^{2} )} : - \frac{4 ( a - b )}{a + b}

x = - \frac{4 ( a - b )}{a + b}; a^{2} + 2 ab + b^{2} \neq = 0

二次方程组的解是:

x = - \frac{4 ( a - b )}{a + b}; a^{2} + 2 ab + b^{2} \neq = 0

5 x (3 x + 2 x) - 3 x = 40

so l v e f or x, (p - 3) \cdot x^{2} - 2 (p - 3) x + 2 = 0

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

so l v e f or x, x^{2} - (t - 1) x + t - 2 = 0

x^{2} - (7 + 3) \cdot x + 6 (1 + 3) = 0