\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2016}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2016}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=0+5=5\\3y=0-4=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
