ĐK: \(x\ne\left\{2,\frac{1}{3},\pm1\right\}\)
\(PT\Leftrightarrow\frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{2x^2-2}=\frac{6}{3x-1}\)
\(\Leftrightarrow\frac{2x-1}{x^2-1}=\frac{6}{3x-1}\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=6\left(x^2-1\right)\)
\(\Leftrightarrow-5x+7=0\Leftrightarrow x=\frac{7}{5}\)
Vậy........
ĐK: x≠{2,13,±1}x≠{2,13,±1}
PT⇔2x+x−22x−1=63x−1PT⇔2x+x−22x−1=63x−1
⇔2(2x−1)2x2−2=63x−1⇔2(2x−1)2x2−2=63x−1
⇔2x−1x2−1=63x−1⇔2x−1x2−1=63x−1
⇔(2x−1)(3x−1)=6(x2−1)⇔(2x−1)(3x−1)=6(x2−1)
⇔−5x+7=0⇔x=7
