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受欢迎的代数 >

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

解答

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

解答

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

求解步骤

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

使用求根公式求解

x_{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

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

将解分隔开

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

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

x = \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:

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

作图

流行的例子

(x+1)^3-(x-2)^3=27

(x + 1)^{3} - (x - 2)^{3} = 27

10 - 3 x^{2} = - 2

4 x^{2} - 20 x = - 29

5 (x - 9)^{2} = 20

solvefor y,x^3+4x^2+xy^2-4y^2=0

so l v e f or y, x^{3} + 4 x^{2} + x y^{2} - 4 y^{2} = 0