Ta có: \(x^2-2x+2^{2y}-2^{y+3}+17=0\)
=>\(x^2-2x+1+\left(2^{y}\right)^2-8\cdot2^{y}+16=0\)
=>\(\left(x-1\right)^2+\left(2^{y}-4\right)^2=0\)
=>\(\begin{cases}x-1=0\\ 2^{y}-4=0\end{cases}\Rightarrow\begin{cases}x=1\\ 2^{y}=4\end{cases}\Rightarrow\begin{cases}x=1\\ y=2\end{cases}\)


