Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2012}\ge0\\\left(3y+4\right)^{2014}\ge0\end{matrix}\right.\forall xy.\)
=> \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\ge0\) \(\forall xy\)
Mà \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0.\)
=> \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}=0\)
=> \(\left(2x-5\right)+\left(3y+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc em học tốt!
