a) Có : \(\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{x+\frac{1}{\frac{x-2+x+1}{x-2}}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{x+\frac{1}{\frac{2x-1}{x-2}}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{\frac{\left(2x-1\right).x+x-2}{2x-1}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{\frac{2x^2-x+x-2}{2x-1}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2}{\frac{2x^2-2}{2x-1}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\frac{2.\left(2x-1\right)}{2x^2-2}=\frac{6}{3x-1}\) \(\Leftrightarrow\) \(\frac{4x-2}{2x^2-2}=\frac{6}{3x-1}\)
\(\Leftrightarrow\) \(\left(4x-2\right)\left(3x-1\right)=6\left(2x^2-2\right)\)
\(\Leftrightarrow\) \(12x^2-4x-6x+2=12x^2-12\)
\(\Leftrightarrow\) \(12x^2-10x+2-12x^2+12=0\)
\(\Leftrightarrow\) \(-10x+14=0\)
\(\Leftrightarrow\) \(-10x=-14\)
\(\Leftrightarrow\) \(x=\frac{-14}{-10}=\frac{14}{10}=\frac{7}{5}\)
Vậy \(x=\frac{7}{5}\)