\(\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left|\left(x-2\right)^{2019}\right|\ge0\\\left(y-1\right)^{2020}\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\ge0\) \(\forall x,y.\)
Mà \(\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}\le0.\)
\(\Rightarrow\left|\left(x-2\right)^{2019}\right|+\left(y-1\right)^{2020}=0\)
\(\Rightarrow\left(x-2\right)^{2019}+\left(y-1\right)^{2020}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^{2019}=0\\\left(y-1\right)^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0+2\\y=0+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{2;1\right\}.\)
Chúc bạn học tốt!
