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受欢迎的代数 >

solvefor a,a^2-2ab+b^2=1

解答

求解

a, a^{2} - 2 ab + b^{2} = 1

解答

a = b + 1, a = b - 1

求解步骤

a^{2} - 2 ab + b^{2} = 1

将

1

para o lado esquerdo

a^{2} - 2 ab + b^{2} - 1 = 0

改写成标准形式

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

a^{2} - 2 ba + b^{2} - 1 = 0

使用求根公式求解

a_{1, 2} = \frac{- ( - 2 b ) \pm ( - 2 b ) ^{2} - 4 \cdot 1 \cdot ( b ^{2} - 1 )}{2 \cdot 1}

(- 2 b)^{2} - 4 \cdot 1 \cdot (b^{2} - 1) = 2

a_{1, 2} = \frac{- ( - 2 b ) \pm 2}{2 \cdot 1}

将解分隔开

a_{1} = \frac{- ( - 2 b ) + 2}{2 \cdot 1}, a_{2} = \frac{- ( - 2 b ) - 2}{2 \cdot 1}

a = \frac{- ( - 2 b ) + 2}{2 \cdot 1} : b + 1

a = \frac{- ( - 2 b ) - 2}{2 \cdot 1} : b - 1

二次方程组的解是:

a = b + 1, a = b - 1

作图

流行的例子

2 g^{2} + 8 g + 8 = 0

x^{2} - 10 = - 6 x

x^{2} - 10 = - 10

solvefor r,v=pir^2

so l v e f or r, v = π r^{2}

s^{2} - 8 s + 16 = 0