Ta có: \(\left|x+2\right|\ge0\forall x\)
\(\left(5y-4\right)^{2018}\ge0\forall y\)
Do đó: \(\left|x+2\right|+\left(5y-4\right)^{2018}\ge0\forall x,y\)
mà \(\left|x+2\right|+\left(5y-4\right)^{2018}=0\)
nên \(\left\{{}\begin{matrix}\left|x+2\right|=0\\\left(5y-4\right)^{2018}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\5y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\5y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\frac{4}{5}\end{matrix}\right.\)
Vậy: x=-2 và \(y=\frac{4}{5}\)
Ta có: \(\left|x+2\right|\ge0\) ; \(\left(5y-4\right)^{2018}\ge0\)
\(\Rightarrow\left|x+2\right|+\left(5y-4\right)^{2018}\ge0\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x+2=0\\5y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\frac{4}{5}\end{matrix}\right.\)
