\(\left(x+1\right)^{10}+\left|y-2\right|^{2019}=0\)
Ta có:
\(\left\{{}\begin{matrix}\left(x+1\right)^{10}\ge0\\\left|y-2\right|^{2019}\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(x+1\right)^{10}+\left|y-2\right|^{2019}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^{10}=0\\\left|y-2\right|^{2019}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+1=0\\\left|y-2\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-1\\y-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{-1;2\right\}.\)
Chúc bạn học tốt!
