\(3x^2+4y^2=6x+13\)
\(\Leftrightarrow3x^2-6x+3+4y^2=16\)
\(\Leftrightarrow3\left(x^2-2x+1\right)+4y^2=16\)
\(\Leftrightarrow3\left(x-1\right)^2+\left(2y\right)^2=16\)
Ta có : \(0\le\left(2y\right)^2\le16\)
\(\Rightarrow\left(2y\right)^2\in\left\{0;1;4;9;16\right\}\)
\(\Rightarrow2y\in\left\{0;1;2;3;4\right\}\)
Mà y nguyên nên \(y\in\left\{0;1;2\right\}\)
+) Với \(y=0\Leftrightarrow3\left(x-1\right)^2=16\Leftrightarrow\left(x-1\right)^2=\frac{16}{3}\)( loại vì x nguyên )
+) Với \(y=1\Leftrightarrow3\left(x-1\right)^2=12\Leftrightarrow\left(x-1\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
+) Với \(y=2\Leftrightarrow3\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy pt có nghiệm \(\left(x;y\right)=\left\{\left(3;1\right);\left(-1;1\right);\left(1;2\right)\right\}\)
