Ta có:\(x^2+x=y^2+y\Rightarrow x^2+x-y^2-y=0\Rightarrow\left(x+y+1\right)\left(x-y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y+1=0\end{matrix}\right.\)
TH1:x-y=0\(\Rightarrow x=y\),khi đó ta có
\(x^2+y^2=2x^2=5\Rightarrow x^2=\frac{5}{2}\Rightarrow x\in\left\{\sqrt{\frac{5}{2}};-\sqrt{\frac{5}{2}}\right\}\)
\(\Rightarrow\left(x,y\right)\in\left\{\left(\sqrt{\frac{5}{2}};\sqrt{\frac{5}{2}}\right),\left(-\sqrt{\frac{5}{2}};-\sqrt{\frac{5}{2}}\right)\right\}\)
TH2:x+y+1=0\(\Rightarrow x=-\left(y+1\right)\),khi đó ta có:
\(x^2+y^2=\left(y+1\right)^2+y^2=5\Rightarrow y^2+2y+1+y^2-5=0\)
\(\Rightarrow2y^2+2y-4=0\Rightarrow y^2+y-2=0\Rightarrow\left(y^2-1\right)+\left(y-1\right)=0\)
\(\Rightarrow\left(y-1\right)\left(y+1\right)+\left(y-1\right)=0\Rightarrow\left(y-1\right)\left(y+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y-1=0\Rightarrow y=1\Rightarrow x=-2\\y+2=0\Rightarrow y=-2\Rightarrow x=1\end{matrix}\right.\)
\(\Rightarrow\left(x,y\right)\in\left\{\left(-2;1\right),\left(1;-2\right)\right\}\)
Vậy \(\left(x,y\right)\in\left\{\left(-\sqrt{\frac{5}{2}};-\sqrt{\frac{5}{2}}\right),\left(\sqrt{\frac{5}{2}};\sqrt{\frac{5}{2}}\right),\left(-2;1\right),\left(1;-2\right)\right\}\)
