c) \(\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(x^2-1\right)^{10}\ge0\\\left(y^2-36\right)^{12}\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\ge0\) \(\forall x,y.\)
Mà \(\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}\le0\)
\(\Rightarrow\left(x^2-1\right)^{10}+\left(y^2-36\right)^{12}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2-1\right)^{10}=0\\\left(y^2-36\right)^{12}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2-1=0\\y^2-36=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=36\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=6\\y=-6\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{1;6\right\},\left\{-1;-6\right\}.\)
Chúc bạn học tốt!
