Đặt $ \dfrac{1}{2x-y} =a ; \dfrac{1}{x+y} = b$ ta có hệ tương đương
$\begin{cases}\\3a-6b =-1\\\\\\a-b=0 \\\\\end{cases}$
$\to$
$\begin{cases}\\3a-6b =-1\\\\\\a=b \\\\\end{cases}$
$\to$
$\begin{cases}\\3a-6a =-1\\\\\\a=b \\\\\end{cases}$
$\to$
$\begin{cases}\\-3a =-1\\\\\\a=b \\\\\end{cases}$
$\to$
$\begin{cases}\\a = \dfrac{1}{3}\\\\\\b = \dfrac{1}{3} \\\\\end{cases}$
$\to$
$\begin{cases}\\ \dfrac{1}{2x-y}= \dfrac{1}{3}\\\\\\ \dfrac{1}{x+y} = \dfrac{1}{3} \\\\\end{cases}$
$\to$
$\begin{cases}\\ 2x-y= 3\\\\\\ x+y = 3 \\\\\end{cases}$
$\to$
$\begin{cases}\\ 3x=6 \\\\\\ x+y = 3 \\\\\end{cases}$
$\to$
$\begin{cases}\\ x =2 \\\\\\ y = 0 \\\\\end{cases}$
Vậy $(x;y) = (2;0)$
