\(\frac{x^2+y^2}{10}=\frac{x^2+2y}{7}\)
\(\Leftrightarrow7\left(x^2+y^2\right)=10\left(x^2-2y^2\right)\)
\(\Leftrightarrow-3x^2+27y^2=0\)
\(\Leftrightarrow-x^2+9y^2=0\)
\(\Leftrightarrow x^2=9y^2\)
\(x^4.y^4=81\Leftrightarrow x^2.y^2=9\Leftrightarrow9y^2.y^2=9\Leftrightarrow y^4=1\)
\(\Rightarrow y=\pm1=>x=\pm1\)
Vậy \(\left(x;y\right)=\left(1,1\right);\left(1;-1\right);\left(-1;-1\right);\left(-1;1\right)\)