Đáp án đúng: C
Giải chi tiết:\(\begin{align} {{\log }_{\sqrt{3}}}\frac{x+y}{{{x}^{2}}+{{y}^{2}}+xy+2}=x(x-3)+y(y-3)+xy\,\,\,\,\,\,\,\,\,\,\,\,(1) \\ \Leftrightarrow {{\log }_{\sqrt{3}}}(x+y)-{{\log }_{\sqrt{3}}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)={{x}^{2}}-3x+{{y}^{2}}-3y+xy \\ \Leftrightarrow {{\log }_{\sqrt{3}}}(x+y)+3x+3y={{\log }_{\sqrt{3}}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)+{{x}^{2}}+{{y}^{2}}+xy \\ \Leftrightarrow {{\log }_{\sqrt{3}}}(x+y)+2+3x+3y={{\log }_{\sqrt{3}}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)+{{x}^{2}}+{{y}^{2}}+xy+2 \\ \Leftrightarrow {{\log }_{\sqrt{3}}}(3x+3y)+3x+3y={{\log }_{\sqrt{3}}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)+{{x}^{2}}+{{y}^{2}}+xy+2\,\,\,\,\,\,\,\,(2) \\ \end{align}\)
Đặt \(f(t)={{\log }_{\sqrt{3}}}t+t,\,\,\,t>0\Rightarrow f'(t)=\frac{1}{t\ln \sqrt{3}}+1>0,\,\,\forall t>0\Rightarrow f(t)\) đồng biến trên \((0;+\infty )\).
\(\begin{align} \left( 2 \right)\Leftrightarrow f(3x+3y)=f({{x}^{2}}+{{y}^{2}}+xy+2)\Leftrightarrow 3x+3y={{x}^{2}}+{{y}^{2}}+xy+2 \\ \Leftrightarrow 4{{x}^{2}}+4{{y}^{2}}+4xy-12x-12y+8=0 \\ \Leftrightarrow {{(2x+y)}^{2}}-6(2x+y)+5=-3{{(y-1)}^{2}}\le 0\Leftrightarrow 1\le 2x+y\le 5 \\ \end{align}\)
Khi đó, \(P=\frac{3x+2y+1}{x+y+6}=1+\frac{2x+y-5}{x+y+6}\le 1\) , vì \(\left\{ \begin{align} 2x+y-5\le 0 \\ x+y+6>0 \\ \end{align} \right.\) .
Vậy, \({{P}_{\max }}=1\) khi và chỉ khi \(\left\{ \begin{array}{l}2x + y - 5 = 0\\y - 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = 1\end{array} \right.\)
Chọn: C
