solvefor x,f^{-1}(x)=2x+1

解答

求解

x, f^{- 1} (x) = 2 x + 1

x = \frac{f}{1 - 2 f}; f \neq = 0, f \neq = \frac{1}{2}

求解步骤

f^{- 1} (x) = 2 x + 1

展开

f^{- 1} (x) : \frac{x}{f}

\frac{x}{f} = 2 x + 1

在两边乘以

f

x = 2 x f + f; f \neq = 0

将

2 x f

para o lado esquerdo

x - 2 x f = f; f \neq = 0

分解

x - 2 x f : x (1 - 2 f)

x (1 - 2 f) = f; f \neq = 0

两边除以

1 - 2 f; f \neq = 0, f \neq = \frac{1}{2}

x = \frac{f}{1 - 2 f}; f \neq = 0, f \neq = \frac{1}{2}

\frac{n}{12} + \frac{n}{3} + \frac{n}{4} = \frac{1}{3}

- 3 \cdot (- 2 - 5 x) - 4 \cdot (x + 6) - (12 x - 2) = 0

3 (x - 5) = \frac{2 x}{4} + \frac{3 ( 1 - 2 x )}{6}

so l v e f or a, a - 4 b + 2 c = 16

so l v e f or a, a + b = 13