Lời giải:
Ta thấy:
\(|x+5|\geq 0, \forall x\in\mathbb{R}\)
\((3y-4)^{2010}=[(3y-4)^{1005}]^2\geq 0, \forall y\in\mathbb{R}\)
Do đó: \(|x+5|+(3y-4)^{2010}\geq 0, \forall x,y\in\mathbb{R}\)
Để dấu "=" xảy ra (theo đề bài) thì \(\left\{\begin{matrix} x+5=0\\ 3y-4=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-5\\ y=\frac{4}{3}\end{matrix}\right.\)
\(\left|x+5\right|+\left(3y-4\right)^{2010}=0\)
Vì \(\left|x+5\right|\ge0\forall x\)
Vì \(\left(3y-4\right)^{2010}\ge0\forall y\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2010}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+5=0\\3y-4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{4}{3}\end{matrix}\right.\)
