\(\frac{2x^2+1}{3x}+\frac{x}{2x-1}=\frac{7x-1}{6}\left(x\ne0;x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow2\left(2x^2+1\right)\left(2x-1\right)+6x^2=x\left(2x-1\right)\left(7x-1\right)\)
\(\Leftrightarrow6x^3-11x^2-3x+2=0\)
\(\Leftrightarrow\left(6x^2+x-1\right)\left(x-2\right)=0\)
Vì 6x2+x-1 \(\ge\)0
=> x-2=0
<=> x=2 (tmđk)