Question

Tìm các số nguyên x, y thỏa mãn:

\(x^2+y^2+5x^2y^2+60=37xy\)

Answer 1

\(PT\Leftrightarrow x^2-2xy+y^2=35xy-5x^2y^2-60\)

\(\Leftrightarrow\left(x-y\right)^2=5\left(3-xy\right)\left(xy-4\right)\)

Mà \(\left(x-y\right)^2\ge0\forall x;y\) nên \(5\left(3-xy\right)\left(xy-4\right)\ge0\Leftrightarrow3\le xy\le4\) 

\(\Rightarrow\hept{\begin{cases}x;y\in\left\{3;4\right\}\\x=y\end{cases}}\) \(\Rightarrow\left(x;y\right)\in\left\{\left(2;2\right);\left(-2;-2\right)\right\}\)

Answer 2

Vi ét à bạn?