`(x+1)/(x-1)-(3x+1)/(x^2-x)=1/x` (Đk :`x ne 0 | xne 1`)

`((x+1)*x)/(x^2-x)-(3x+1)/(x^2-x)=(x-1)/(x^2-x)`

`(x+1)x -(3x+1)=x-1`

`x^2 +x -3x -1=x-1`

`x^2 +x-3x-x=-1+1`

`x^2-3x=0`

`x(x-3)=0`

$\left[\begin{matrix} x=0\\ x-3=0\end{matrix}\right.$

$\left[\begin{matrix} x=0 (KTM) \\ x=3(TM) \end{matrix}\right.$

Vậy `x=3`