(k-1)y^2-5y+3k-7=0

解答

(k - 1) y^{2} - 5 y + 3 k - 7 = 0

y = \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, y = \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

求解步骤

(k - 1) y^{2} - 5 y + 3 k - 7 = 0

使用求根公式求解

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( k - 1 ) ( 3 k - 7 )}{2 ( k - 1 )}

化简

(- 5)^{2} - 4 (k - 1) (3 k - 7) : - 12 k^{2} + 40 k - 3

y_{1, 2} = \frac{- ( - 5 ) \pm - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

将解分隔开

y_{1} = \frac{- ( - 5 ) + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, y_{2} = \frac{- ( - 5 ) - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

y = \frac{- ( - 5 ) + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )} : \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

y = \frac{- ( - 5 ) - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )} : \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}

二次方程组的解是:

y = \frac{5 + - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}, y = \frac{5 - - 12 k ^{2} + 40 k - 3}{2 ( k - 1 )}; k \neq = 1

- 2 x^{2} - 6 x + 7 = 0

2 u^{2} - 5 u - 3 = 0

so l v e f or x, y = - 0.02 x^{2} + x + 2.6

4 x (x - 2) + 3 (x - 3) = - 3

so l v e f or x, y^{4} = (x + 2)^{2}