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受欢迎的代数 >

solvefor t,st+t=t^2+s

解答

求解

t, s t + t = t^{2} + s

解答

t = s, t = 1

求解步骤

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

交换两边

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

将

t

para o lado esquerdo

t^{2} + s - t = s t

将

s t

para o lado esquerdo

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

改写成标准形式

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

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

使用求根公式求解

t_{1, 2} = \frac{- ( - 1 - s ) \pm ( - 1 - s ) ^{2} - 4 \cdot 1 \cdot s}{2 \cdot 1}

化简

(- 1 - s)^{2} - 4 \cdot 1 \cdot s : s - 1

t_{1, 2} = \frac{- ( - 1 - s ) \pm ( s - 1 )}{2 \cdot 1}

将解分隔开

t_{1} = \frac{- ( - 1 - s ) + s - 1}{2 \cdot 1}, t_{2} = \frac{- ( - 1 - s ) - ( s - 1 )}{2 \cdot 1}

t = \frac{- ( - 1 - s ) + s - 1}{2 \cdot 1} : s

t = \frac{- ( - 1 - s ) - ( s - 1 )}{2 \cdot 1} : 1

二次方程组的解是:

t = s, t = 1

作图

流行的例子

y^{2} - 18 y + 45 = 0

1.81L^2+22.275L-750=0

1.81 L^{2} + 22.275 L - 750 = 0

x^{2} = \frac{- 5}{2}

d^{2} - 1 = 0

2 x^{2} - 10 x + 2 = 0