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受欢迎的代数 >

solvefor u,v(2uv^2-3)du+(3u^2v^2-3+4v)dv=0

解答

求解

u, v (2 u v^{2} - 3) d u + (3 u^{2} v^{2} - 3 + 4 v) d v = 0

解答

u = \frac{3 + - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}, u = \frac{3 - - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}; 5 d v^{3} \neq = 0

求解步骤

v (2 u v^{2} - 3) d u + (3 u^{2} v^{2} - 3 + 4 v) d v = 0

展开

v (2 u v^{2} - 3) d u + (3 u^{2} v^{2} - 3 + 4 v) d v : 5 d u^{2} v^{3} + 4 d v^{2} - 3 d v - 3 d uv

5 d u^{2} v^{3} + 4 d v^{2} - 3 d v - 3 d uv = 0

改写成标准形式

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

5 d v^{3} u^{2} - 3 d vu + 4 d v^{2} - 3 d v = 0

使用求根公式求解

u_{1, 2} = \frac{- ( - 3 d v ) \pm ( - 3 d v ) ^{2} - 4 \cdot 5 d v ^{3} ( 4 d v ^{2} - 3 d v )}{2 \cdot 5 d v ^{3}}

化简

(- 3 d v)^{2} - 4 \cdot 5 d v^{3} (4 d v^{2} - 3 d v) : d v 9 - 20 v^{2} (- 3 + 4 v)

u_{1, 2} = \frac{- ( - 3 d v ) \pm d v 9 - 20 v ^{2} ( - 3 + 4 v )}{2 \cdot 5 d v ^{3}}; 5 d v^{3} \neq = 0

将解分隔开

u_{1} = \frac{- ( - 3 d v ) + d v 9 - 20 v ^{2} ( - 3 + 4 v )}{2 \cdot 5 d v ^{3}}, u_{2} = \frac{- ( - 3 d v ) - d v 9 - 20 v ^{2} ( - 3 + 4 v )}{2 \cdot 5 d v ^{3}}

u = \frac{- ( - 3 d v ) + d v 9 - 20 v ^{2} ( - 3 + 4 v )}{2 \cdot 5 d v ^{3}} : \frac{3 + - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}

u = \frac{- ( - 3 d v ) - d v 9 - 20 v ^{2} ( - 3 + 4 v )}{2 \cdot 5 d v ^{3}} : \frac{3 - - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}

二次方程组的解是:

u = \frac{3 + - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}, u = \frac{3 - - 20 v ^{2} ( 4 v - 3 ) + 9}{10 v ^{2}}; 5 d v^{3} \neq = 0

作图

流行的例子

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

\frac{x ^{2} - x + 5}{4} = 0

- 63 = - 2 y^{2} + 9

(x+1)(x+2)-(x-3)(+4)=6

(x + 1) (x + 2) - (x - 3) (+ 4) = 6

x^{2} + x + 1 = 7