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受欢迎的代数 >

solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0

解答

求解

s, t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t = 0

解答

s = 0, s = - t; 2 d t \neq = 0

求解步骤

t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t = 0

展开

t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t : 2 d s t^{2} + 2 d s^{2} t

2 d s t^{2} + 2 d s^{2} t = 0

改写成标准形式

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

2 d t s^{2} + 2 d t^{2} s = 0

使用求根公式求解

s_{1, 2} = \frac{- 2 d t ^{2} \pm ( 2 d t ^{2} ) ^{2} - 4 \cdot 2 d t \cdot 0}{2 \cdot 2 d t}

化简

(2 d t^{2})^{2} - 4 \cdot 2 d t \cdot 0 : 2 d t^{2}

s_{1, 2} = \frac{- 2 d t ^{2} \pm 2 d t ^{2}}{2 \cdot 2 d t}; 2 d t \neq = 0

将解分隔开

s_{1} = \frac{- 2 d t ^{2} + 2 d t ^{2}}{2 \cdot 2 d t}, s_{2} = \frac{- 2 d t ^{2} - 2 d t ^{2}}{2 \cdot 2 d t}

s = \frac{- 2 d t ^{2} + 2 d t ^{2}}{2 \cdot 2 d t} : 0

s = \frac{- 2 d t ^{2} - 2 d t ^{2}}{2 \cdot 2 d t} : - t

二次方程组的解是:

s = 0, s = - t; 2 d t \neq = 0

作图

流行的例子

solvefor x,x^2+y^2+2x+2y+34=0

so l v e f or x, x^{2} + y^{2} + 2 x + 2 y + 34 = 0

(3x-2)(x+3)+(x-2)(x+5)=-2(x+8)

(3 x - 2) (x + 3) + (x - 2) (x + 5) = - 2 (x + 8)

32 = - 16 t^{2} + 48 t

24 x^{2} + 36 x - 24 = 0

因式分解 3x^2+5x+2=0

f a c t or 3 x^{2} + 5 x + 2 = 0