Question

Chứng minh đẳng thức
\(\text{[}\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\text{]}:\frac{x-1}{x}=\frac{2x}{x-1}\)

Answer 1

ĐKXĐ:...

\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}=\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{-3x^2-2x+1}{3x}\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)\left(1-3x\right)}{3x\left(x+1\right)}\right].\frac{x}{x-1}=\left(\frac{2}{3x}-\frac{2\left(1-3x\right)}{3x}\right).\left(\frac{x}{x-1}\right)\)

\(=\left(\frac{2-2+6x}{3x}\right)\left(\frac{x}{x-1}\right)=\frac{2x}{x-1}\)