Bài giải:

$\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}$

ĐKXĐ: x $\ne $2; x $\ne $$\frac{1}{3}$ ; x $\ne $-1; x $\ne $1.

$\Leftrightarrow \frac{2}{x+\frac{1}{\frac{x-2+x+1}{x-2}}}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2}{x+\frac{1}{\frac{2x-1}{x-2}}}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2}{\frac{2{{x}^{2}}-x+x-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2}{\frac{2{{x}^{2}}-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow 2:\frac{2{{x}^{2}}-2}{2x-1}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2.(2x-1)}{2{{x}^{2}}-2}=\frac{6}{3x-1}$

$\Leftrightarrow \frac{2x-1}{{{x}^{2}}-1}=\frac{6}{3x-1}$

$\Leftrightarrow $(2x – 1)(3x – 1) = 6(x² – 1)

$\Leftrightarrow $6x² – 5x + 1 = 6x² – 6

$\Leftrightarrow $-5x = -7

$\Leftrightarrow $x = $\frac{7}{5}$      (thỏa mãn)

Vậy x = $\frac{7}{5}$