Xét :
\(\left(2x-3\right)^{2012}\ge0\) ( với mọi giá trị x )
\(\left(5y+2\right)^{2014}\ge0\) ( với mọi giá trị y )
\(\Rightarrow\left(2x+3\right)^{2012}+\left(5y+2\right)^{2014}\ge0\) ( nghịch lí với đề bài )
Ta có: \(\left(2x-3\right)^{2012}=\left[\left(2x-3\right)^{1006}\right]^2\ge0\forall x\)
\(\left(5y+2\right)^{2014}=\left[\left(5y+2\right)^{1007}\right]^2\ge0\forall x\)
\(\Leftrightarrow\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}\ge0\forall x\)
mà theo đề có: \(\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}\le0\)
\(\Rightarrow\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^{2012}=0\\\left(5y+2\right)^{2014}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\5y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy ...
Vì \(\left(2x-3\right)^{2012}\ge0\forall x\)
\(\left(5y+2\right)^{2014}\ge0\forall y\)
\(\Rightarrow\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}\ge0\)
mà theo đề có:
\(\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}\le0\)
\(\Rightarrow\left(2x-3\right)^{2012}+\left(5y+2\right)^{2014}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^{2012}=0\\\left(5y+2\right)^{2014}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\5y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
