Ta có \(2^x+2^y=2^{x+y}\)
⇔ \(2^x+2^y-2^{x.y}\) = 0
⇔ \(2^x+2^y-2^x.2^y-1=-1\)
⇔ \(2^x\left(1+2^y\right)-\left(1+2^y\right)=-1\)
⇔ \(\left(1+2^y\right)\left(2^x-1\right)=-1\)
⇔ \(\left\{{}\begin{matrix}1+2^y=1\\2^x-1=-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}1-2^y=-1\\2^x-1=1\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2^y=0\\2^x=0\end{matrix}\right.\) ( vô lí ) hoặc \(\left\{{}\begin{matrix}2^y=2\\2^x=2\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\) ( thỏa mãn x , y ∈ Z )
Vậy \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) thỏa mãn đề bài
@@ Học tốt @@
## Chiyuki Fujito
