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受欢迎的代数 >

solvefor y,5x^3y^2-2x^2=9xy+2

解答

求解

y, 5 x^{3} y^{2} - 2 x^{2} = 9 x y + 2

解答

y = \frac{9 + 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}, y = \frac{9 - 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}; x \neq = 0

求解步骤

5 x^{3} y^{2} - 2 x^{2} = 9 x y + 2

将

2

para o lado esquerdo

5 x^{3} y^{2} - 2 x^{2} - 2 = 9 x y

将

9 x y

para o lado esquerdo

5 x^{3} y^{2} - 2 x^{2} - 2 - 9 x y = 0

改写成标准形式

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

5 x^{3} y^{2} - 9 x y - 2 x^{2} - 2 = 0

使用求根公式求解

y_{1, 2} = \frac{- ( - 9 x ) \pm ( - 9 x ) ^{2} - 4 \cdot 5 x ^{3} ( - 2 x ^{2} - 2 )}{2 \cdot 5 x ^{3}}

化简

(- 9 x)^{2} - 4 \cdot 5 x^{3} (- 2 x^{2} - 2) : x 81 + 40 x (1 + x^{2})

y_{1, 2} = \frac{- ( - 9 x ) \pm x 81 + 40 x ( 1 + x ^{2} )}{2 \cdot 5 x ^{3}}; x \neq = 0

将解分隔开

y_{1} = \frac{- ( - 9 x ) + x 81 + 40 x ( 1 + x ^{2} )}{2 \cdot 5 x ^{3}}, y_{2} = \frac{- ( - 9 x ) - x 81 + 40 x ( 1 + x ^{2} )}{2 \cdot 5 x ^{3}}

y = \frac{- ( - 9 x ) + x 81 + 40 x ( 1 + x ^{2} )}{2 \cdot 5 x ^{3}} : \frac{9 + 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}

y = \frac{- ( - 9 x ) - x 81 + 40 x ( 1 + x ^{2} )}{2 \cdot 5 x ^{3}} : \frac{9 - 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}

二次方程组的解是:

y = \frac{9 + 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}, y = \frac{9 - 40 x ( x ^{2} + 1 ) + 81}{10 x ^{2}}; x \neq = 0

作图

流行的例子

solvefor x,x^2=2t^2+6t

so l v e f or x, x^{2} = 2 t^{2} + 6 t

solvefor h,h^2+2k-3=0

so l v e f or h, h^{2} + 2 k - 3 = 0

solvefor x,x^2+y^3-12xy=0

so l v e f or x, x^{2} + y^{3} - 12 x y = 0

(x)=x^2-40x+500

(x) = x^{2} - 40 x + 500

(3x+6)(x+6)=231

(3 x + 6) (x + 6) = 231