a) Ta có \(x^5+y^5=x^2+y^2\Leftrightarrow x^5-x^2+y^5-y^2=0\Leftrightarrow x^2\left(x^3-1\right)+y^2\left(y^3-1\right)=0\Leftrightarrow x^2\left(x^3-x^3-y^3\right)+y^2\left(y^3-y^3-x^3\right)=0\Leftrightarrow-x^2.y^3-y^2.x^3=0\Leftrightarrow-x^2.y^2\left(y+x\right)=0\Leftrightarrow x^2y^2\left(x+y\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x^2.y^2=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\x=-y\end{matrix}\right.\)
Trường hợp 1:
\(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Trường hợp 2:x=-y
Ta có \(x^3+y^3=1\Leftrightarrow-y^3+y^3=1\Leftrightarrow0=1\left(ktm\right)\)
Vậy (x;y)={(0;1);(1;0)}
b) Ta có \(x^3+y^3=x^2+y^2\Leftrightarrow x^3-x^2+y^3-y^2=0\Leftrightarrow x^2\left(x-1\right)+y^2\left(y-1\right)=0\Leftrightarrow x^2\left(x-x-y\right)+y^2\left(y-y-x\right)=0\Leftrightarrow-x^2y-y^2x=0\Leftrightarrow-xy\left(x+y\right)=0\Leftrightarrow xy\left(x+y\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}xy=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\x+y=0\end{matrix}\right.\)\(\)
Trường hợp 1:
\(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Trường hợp 2:
x+y=0 mà x+y=1 nên ktm
Vậy (x;y)={(0;1);(1;0)}
