Cho $x,y$ là các số thực dương thỏa mãn ${{\log...

${{\log }_{2}}\dfrac{3x+3y+4}{{{x}^{2}}+{{y}^{2}}}=\left( x+y-1 \right)\left( 2x+2y-1 \right)-4\left( xy+1 \right)$.
$\Leftrightarrow {{\log }_{2}}\left( 3x+3y+4 \right)-{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)-\left( 3x+3y+4 \right)+1$.
$\Leftrightarrow {{\log }_{2}}\left( 3x+3y+4 \right)-\left[ {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)+1 \right]=2\left( {{x}^{2}}+{{y}^{2}} \right)-\left( 3x+3y+4 \right)$.
$\Leftrightarrow {{\log }_{2}}\left( 3x+3y+4 \right)-{{\log }_{2}}2\left( {{x}^{2}}+{{y}^{2}} \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)-\left( 3x+3y+4 \right)$.
$\Leftrightarrow \left( 3x+3y+4 \right)+{{\log }_{2}}\left( 3x+3y+4 \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)+{{\log }_{2}}2\left( {{x}^{2}}+{{y}^{2}} \right)\ \ \left( * \right)$
Xét hàm số $f\left( t \right)=t+{{\log }_{2}}t$ đồng biến trên khoảng $\left( 0;+\infty \right)$ nên $\left( * \right)\Leftrightarrow \left( 3x+3y+4 \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)\ \ $
Ta có ${{\left( x+y \right)}^{2}}\le 2\left( {{x}^{2}}+{{y}^{2}} \right)\Leftrightarrow {{\left( x+y \right)}^{2}}\le \left( 3x+3y+4 \right)\ \Leftrightarrow {{\left( x+y \right)}^{2}}-3\left( x+y \right)-4\le 0\ $
$\Leftrightarrow -1\le x+y\le 4$. Do $x,y$ là các số thực dương nên $0<x+y\le 4\Rightarrow x+y-4\le 0$.
Ta có $\left\{ \begin{aligned}
& x+y-4\le 0 \\
& 2x+y+1>0 \\
\end{aligned} \right.\Rightarrow \dfrac{x+y-4}{2x+y+1}\le 0$
Suy ra $P=\dfrac{5x+3y-2}{2x+y+1}=\dfrac{2\left( 2x+y+1 \right)+\left( x+y-4 \right)}{2x+y+1}=2+\dfrac{x+y-4}{2x+y+1}\le 2$.
Vậy ${{P}_{\max }}=2$ xảy ra khi $x=y=2$.