Lời giải:
Ta có: \(2x^2+5y^2=12\)
\(\Rightarrow 12-5y^2=2x^2\geq 0\)
\(\Rightarrow 5y^2\leq 12\Rightarrow y^2\leq \frac{12}{5}<4\)
\(\Rightarrow -2< y< 2\)
Vì \(y\in\mathbb{Z}\Rightarrow y\in\left\{-1;0;1\right\}\)
Nếu \(y\pm 1\Rightarrow 3x^2=12-5y^2=12-5=7\)
\(\Rightarrow x^2=\frac{7}{3}\) (vô lý do $x$ nguyên)
Nếu \(y=0\Rightarrow 3x^2=12-5y^2=12\Rightarrow x^2=4\Rightarrow x=\pm 2\)
Vậy \((x,y)=(\pm 2,0)\)