xem như pt bậc 2 ẩn x
x^2 + y^2 + 5(xy)^2 + 60 =37xy
<>(1+5y^2).x^2 -37xy + 60 + y^2 =0
denta = 37^2*y^2 - 4*(60+y^2)*(1+5y^2)
= -20y^4+165y^2- 240 >=0
=> 1 < y^2 <7 => y= +-2
với y= 2 => x = 2 thỏa mãn
với y =-2 => x =- 2 thỏa mãn
xong nha
cách khác :
\(x^2+y^2+5x^2y^2+60=37xy\)
\(\left(x^2+y^2-2xy\right)=35xy-5x^2y^2-60\)
\(\left(x-y\right)^2=5\left(4-xy\right)\left(xy-3\right)\)
vì \(\left(x-y\right)^2\ge0\)\(\Rightarrow5\left(4-xy\right)\left(xy-3\right)\ge0\)
\(\Rightarrow\left(4-xy\right)\left(xy-3\right)\ge0\)
\(\Rightarrow3\le xy\le4\)
vì \(x,y\in Z\Rightarrow xy\in Z\)nên \(xy\in\left\{3,4\right\}\)
\(\Rightarrow\left(x-y\right)^2=5\left(4-xy\right)\left(xy-3\right)=0\)
\(\Rightarrow x=y\)
+) \(xy=3\Rightarrow x^2=3\Rightarrow x=\pm\sqrt{3}\left(ktm\right)\)
+) \(xy=4\Rightarrow x^2=4\Rightarrow x=\pm2\)
Vậy \(x=y=\pm2\)
