Ta có 2x+y-2x-2y=0
<=> 2x(2y-1)-(2y-1)=1
<=> (2x-1)(2y-1)=1
TH1
\(\hept{\begin{cases}2^x-1=1\\2^y-1=1\end{cases}}\)<=> \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
TH2
\(\hept{\begin{cases}2^x-1=-1\\2^y-1=-1\end{cases}}\)<=>\(\hept{\begin{cases}x=0\\y=0\end{cases}}\)