Ta có : \(\left\{{}\begin{matrix}\left|5x+1\right|\ge0\forall x\\\left|2y-1\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow3+\left|5x+1\right|+\left|2y-1\right|\ge3\forall x;y\)
\(\Rightarrow\dfrac{12}{3+\left|5x+1\right|+\left|2y-1\right|}\le\dfrac{12}{3}=4\forall x;y\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|5x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+1=0\\2y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=-1\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy Max của b/t trên là : \(4\Leftrightarrow x=-\dfrac{1}{5};y=\dfrac{1}{2}\)
