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受欢迎的代数 >

solvefor y,x^3+5x^2y+2y^2=4y+11

解答

求解

y, x^{3} + 5 x^{2} y + 2 y^{2} = 4 y + 11

解答

y = \frac{- 5 x ^{2} + 4 + 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}, y = \frac{- 5 x ^{2} + 4 - 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}

求解步骤

x^{3} + 5 x^{2} y + 2 y^{2} = 4 y + 11

将

11

para o lado esquerdo

x^{3} + 5 x^{2} y + 2 y^{2} - 11 = 4 y

将

4 y

para o lado esquerdo

x^{3} + 5 x^{2} y + 2 y^{2} - 11 - 4 y = 0

改写成标准形式

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

2 y^{2} + (5 x^{2} - 4) y + x^{3} - 11 = 0

使用求根公式求解

y_{1, 2} = \frac{- ( 5 x ^{2} - 4 ) \pm ( 5 x ^{2} - 4 ) ^{2} - 4 \cdot 2 ( x ^{3} - 11 )}{2 \cdot 2}

化简

(5 x^{2} - 4)^{2} - 4 \cdot 2 (x^{3} - 11) : 25 x^{4} - 8 x^{3} - 40 x^{2} + 104

y_{1, 2} = \frac{- ( 5 x ^{2} - 4 ) \pm 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{2 \cdot 2}

将解分隔开

y_{1} = \frac{- ( 5 x ^{2} - 4 ) + 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{2 \cdot 2}, y_{2} = \frac{- ( 5 x ^{2} - 4 ) - 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{2 \cdot 2}

y = \frac{- ( 5 x ^{2} - 4 ) + 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{2 \cdot 2} : \frac{- 5 x ^{2} + 4 + 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}

y = \frac{- ( 5 x ^{2} - 4 ) - 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{2 \cdot 2} : \frac{- 5 x ^{2} + 4 - 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}

二次方程组的解是:

y = \frac{- 5 x ^{2} + 4 + 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}, y = \frac{- 5 x ^{2} + 4 - 25 x ^{4} - 8 x ^{3} - 40 x ^{2} + 104}{4}

作图

流行的例子

(2 x + 5)^{2} = 25

x (x - 4) = 5 (x - 4)

(x - 3) (x + 2) = 14

(2 x + 5)^{2} = - 4

(x - 4) (2 x - 3) = 12