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受欢迎的代数 >

solvefor t,s_{t}(t)(t)=f(p_{t}(t)(t)-1)

解答

求解

t, s_{t} (t) (t) (t) = f (p_{t} (t) (t) (t) - 1)

解答

t = i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}, t = - i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}; s_{t} (t) \neq = p_{t} (t) f

求解步骤

s_{t} (t) (t) (t) = f (p_{t} (t) (t) (t) - 1)

将

f (p_{t} (t) (t) (t) - 1)

para o lado esquerdo

t^{2} s_{t} (t) - f (t^{2} p_{t} (t) - 1) = 0

展开

t^{2} s_{t} (t) - f (t^{2} p_{t} (t) - 1) : s_{t} (t) t^{2} - p_{t} (t) f t^{2} + f

s_{t} (t) t^{2} - p_{t} (t) f t^{2} + f = 0

将

f

到右边

s_{t} (t) t^{2} - p_{t} (t) f t^{2} = - f

分解

s_{t} (t) t^{2} - p_{t} (t) f t^{2} : t^{2} (s_{t} (t) - p_{t} (t) f)

t^{2} (s_{t} (t) - p_{t} (t) f) = - f

两边除以

s_{t} (t) - p_{t} (t) f; s_{t} (t) \neq = p_{t} (t) f

t^{2} = - \frac{f}{s _{t} ( t ) - p _{t} ( t ) f}; s_{t} (t) \neq = p_{t} (t) f

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

t = - \frac{f}{s _{t} ( t ) - p _{t} ( t ) f}, t = - - \frac{f}{s _{t} ( t ) - p _{t} ( t ) f}; s_{t} (t) \neq = p_{t} (t) f

化简

- \frac{f}{s _{t} ( t ) - p _{t} ( t ) f} : i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}

化简

- - \frac{f}{s _{t} ( t ) - p _{t} ( t ) f} : - i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}

t = i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}, t = - i \frac{f}{s _{t} ( t ) - f p _{t} ( t )}; s_{t} (t) \neq = p_{t} (t) f

作图

流行的例子

solvefor x,-5x+3y=-60

so l v e f or x, - 5 x + 3 y = - 60

solvefor x,p^2-(x+3y)p+2y(x+y)=0

so l v e f or x, p^{2} - (x + 3 y) p + 2 y (x + y) = 0

(-11+u)/6 =-203

\frac{- 11 + u}{6} = - 203

5(x+8)+4(x-6)=61

5 (x + 8) + 4 (x - 6) = 61

3 c + 4 u = 33