\(x^{2013}+y^{2013}=\left(x^3+y^3\right)\left(x^{2013}+y^{2013}\right)=x^{2016}+y^{2016}+x^3y^3\left(x^{2010}+y^{2010}\right)=x^{2013}+y^{2013}+x^3y^3\left(x^{2010}+y^{2010}\right)\)\(\Rightarrow x^3y^3\left(x^{2010}+y^{2010}\right)=0\)
*\(x=0\Rightarrow y=1\); \(y=0\Rightarrow x=1\).
*\(x^{2010}+y^{2010}=0\) mà \(\left\{{}\begin{matrix}x^{2010}\ge0\\y^{2010}\ge0\end{matrix}\right.\)
\(\Rightarrow x=y=0\) trái với giả thiết \(\Rightarrow\)loại.
Vậy các cặp số (x,y) là (0,1) ; (1,0).
\(2x^2+2xy-x-y-3=0\)
\(\Rightarrow2x\left(x+y\right)-\left(x+y\right)=3\)
\(\Rightarrow\left(x+y\right)\left(2x-1\right)=3\)
Ta có: \(x,y\in Z\) và \(3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)
Ta xét mỗi trường hợp:
\(TH_1:\left\{{}\begin{matrix}x+y=1\\2x-1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
\(TH_2:\left\{{}\begin{matrix}x+y=3\\2x-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(TH_3:\left\{{}\begin{matrix}x+y=-1\\2x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
\(TH_4:\left\{{}\begin{matrix}x+y=-3\\2x-1=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\)
Vậy các cặp số (x,y) nguyên thỏa mãn phương trình là: (2,-1) ; (1,2) ; (-1,0) ; (0,-3).
