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受欢迎的代数 >

solvefor t,t=(2a(bt^2+n))/3+d/2

解答

求解

t, t = \frac{2 a ( b t ^{2} + n )}{3} + \frac{d}{2}

解答

t = \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}, t = \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}; 4 ab \neq = 0

求解步骤

t = \frac{2 a ( b t ^{2} + n )}{3} + \frac{d}{2}

找到

3, 2

的最小公倍数:

6

乘以最小公倍数=

6

t \cdot 6 = \frac{2 a ( b t ^{2} + n )}{3} \cdot 6 + \frac{d}{2} \cdot 6

化简

6 t = 4 a (n + t^{2} b) + 3 d

展开

4 a (n + t^{2} b) + 3 d : 4 na + 4 t^{2} ab + 3 d

6 t = 4 na + 4 t^{2} ab + 3 d

交换两边

4 na + 4 t^{2} ab + 3 d = 6 t

将

6 t

para o lado esquerdo

4 na + 4 t^{2} ab + 3 d - 6 t = 0

改写成标准形式

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

4 ab t^{2} - 6 t + 4 na + 3 d = 0

使用求根公式求解

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ab ( 4 na + 3 d )}{2 \cdot 4 ab}

化简

(- 6)^{2} - 4 \cdot 4 ab (4 na + 3 d) : 2 9 - 4 ab (3 d + 4 na)

t_{1, 2} = \frac{- ( - 6 ) \pm 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}; 4 ab \neq = 0

将解分隔开

t_{1} = \frac{- ( - 6 ) + 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}, t_{2} = \frac{- ( - 6 ) - 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab}

t = \frac{- ( - 6 ) + 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab} : \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}

t = \frac{- ( - 6 ) - 2 9 - 4 ab ( 3 d + 4 na )}{2 \cdot 4 ab} : \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}

二次方程组的解是:

t = \frac{3 + - 4 ab ( 4 na + 3 d ) + 9}{4 ab}, t = \frac{3 - - 4 ab ( 4 na + 3 d ) + 9}{4 ab}; 4 ab \neq = 0

作图

流行的例子

solvefor x,x^2y+xy=1

so l v e f or x, x^{2} y + x y = 1

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

5^2=x^2+(x+1)^2

5^{2} = x^{2} + (x + 1)^{2}

12x^2-256x+1024=0

12 x^{2} - 256 x + 1024 = 0

8 x^{2} - 5 = 67