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受欢迎的代数 >

solvefor x,f=4x^2y^2-5xy^3+12

解答

求解

x, f = 4 x^{2} y^{2} - 5 x y^{3} + 12

解答

x = \frac{5 y ^{2} + 25 y ^{4} - 16 ( - f + 12 )}{8 y}, x = \frac{5 y ^{2} - 25 y ^{4} - 16 ( - f + 12 )}{8 y}; y \neq = 0

求解步骤

f = 4 x^{2} y^{2} - 5 x y^{3} + 12

交换两边

4 x^{2} y^{2} - 5 x y^{3} + 12 = f

将

f

para o lado esquerdo

4 x^{2} y^{2} - 5 x y^{3} + 12 - f = 0

改写成标准形式

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

4 y^{2} x^{2} - 5 y^{3} x + 12 - f = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 5 y ^{3} ) \pm ( - 5 y ^{3} ) ^{2} - 4 \cdot 4 y ^{2} ( 12 - f )}{2 \cdot 4 y ^{2}}

化简

(- 5 y^{3})^{2} - 4 \cdot 4 y^{2} (12 - f) : y 25 y^{4} - 16 (12 - f)

x_{1, 2} = \frac{- ( - 5 y ^{3} ) \pm y 25 y ^{4} - 16 ( 12 - f )}{2 \cdot 4 y ^{2}}; y \neq = 0

将解分隔开

x_{1} = \frac{- ( - 5 y ^{3} ) + y 25 y ^{4} - 16 ( 12 - f )}{2 \cdot 4 y ^{2}}, x_{2} = \frac{- ( - 5 y ^{3} ) - y 25 y ^{4} - 16 ( 12 - f )}{2 \cdot 4 y ^{2}}

x = \frac{- ( - 5 y ^{3} ) + y 25 y ^{4} - 16 ( 12 - f )}{2 \cdot 4 y ^{2}} : \frac{5 y ^{2} + 25 y ^{4} - 16 ( - f + 12 )}{8 y}

x = \frac{- ( - 5 y ^{3} ) - y 25 y ^{4} - 16 ( 12 - f )}{2 \cdot 4 y ^{2}} : \frac{5 y ^{2} - 25 y ^{4} - 16 ( - f + 12 )}{8 y}

二次方程组的解是:

x = \frac{5 y ^{2} + 25 y ^{4} - 16 ( - f + 12 )}{8 y}, x = \frac{5 y ^{2} - 25 y ^{4} - 16 ( - f + 12 )}{8 y}; y \neq = 0

作图

流行的例子

5-2(3x^2-2x)=x^2+7-2x^2

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

x^{2} - 6 x = - 12

x^2-12x-5=-6x+2

x^{2} - 12 x - 5 = - 6 x + 2

9 p^{2} - 27 p + 20 = 0

solvefor x,x^2y^2-x-4y^2=0

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