\(x^2+2^y=11\)
\(\Leftrightarrow x^2+2^y=3^2+2^1\)
\(\Rightarrow x=3\)
\(\Rightarrow y=1\)
Ta có : \(x^2+2^y=11\)
\(x^2+2^y=3^2+2^1\)
\(\Leftrightarrow\) \(x^2=3^2\Rightarrow x=3\)
\(\Leftrightarrow2^y=2^1\Rightarrow y=1\)
Vậy \(x=3;y=1\)
x2 + 2y = 11
=> 11 = 32 + 21
=> x2 = 32
=> 2y = 21
Vậy: x = 3; y = 1