$\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2$
ĐK: $\begin{cases}x+1\ne 0\\ x\ne 0\end{cases}↔\begin{cases}x\ne -1\\ x\ne 0\end{cases}$
$↔ \dfrac{x(x+3)+(x-2)(x-1)}{x(x+1)}=2$
$↔ \dfrac{x^2+3x+x^2+x-2x-2}{x(x+1)}=2$
$↔ \dfrac{2x^2+2x-2}{x(x+1)}=\dfrac{2x(x+1)}{x(x+1)}$
$↔ 2x^2+2x-2=2x(x+1)$
$↔ 2x^2+2x-2=2x^2+2x$
$↔ 2x^2+2x-2-2x^2-2x=0$
$↔ -2=0$ (vô lý)
Vậy $x\in \varnothing$
