\(2^x+2^y=496\) \(\Rightarrow\frac{2^y.2^x}{2^y}+2^y=496\) \(\Rightarrow2^y.2^{x-y}+2^y=2^4.31\) \(\Rightarrow2^y\left(2^{x-y}+1\right)=2^4.31\)
Th1: x = y, ta có: \(2^y\left(2^{x-y}+1\right)=2^4.31\)\(\Rightarrow2^y\left(2^{x-x}+1\right)=2^4.31\)\(\Rightarrow2^y\left(2^0+1\right)=2^4.31\)
\(\Rightarrow2^y.2=2^4.31\)\(\Rightarrow2^y=2^3.31\)(Vô lý)
Th2: x ≠ y, ta có:\(2^y⋮2\); \(2^{x-y}⋮2\)\(\Rightarrow2^{x-y}+1\)chia cho 2 dư 1 mà 31 chia cho 2 dư 1
\(\Rightarrow\hept{\begin{cases}2^y=2^4\\2^{x-y}+1=31\end{cases}}\Rightarrow\hept{\begin{cases}y=4\\2^{x-y}=30\end{cases}}\Rightarrow\hept{\begin{cases}y=4\\x\in\varnothing\end{cases}}\)(Vô lý)
Vậy không có trường hợp x, y nào thỏa mãn 2x + 2y = 496
