Có bao nhiêu cặp số nguyên dương $(x ; y)$ thỏa mãn $\log _2...

${{\log }_{2}}\dfrac{x+y}{{{x}^{2}}+{{y}^{2}}+2}\ge x(x-4)+y(y-4)\Leftrightarrow {{\log }_{2}}\left( x+y \right)-{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+2 \right)\ge {{x}^{2}}+{{y}^{2}}-4\left( x+y \right)$
$\Leftrightarrow {{\log }_{2}}4\left( x+y \right)+4\left( x+y \right)\ge {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+2 \right)+{{x}^{2}}+{{y}^{2}}+2.$
Đặt $f\left( t \right)={{\log }_{2}}t+t,\ \left( t>0 \right)\Rightarrow {f}'\left( t \right)=\dfrac{1}{t\ln 2}+1>0,\forall t>0.$
Ta có $\left\{ \begin{aligned}
& {f}'\left( t \right)>0 \\
& f\left( 4\left( x+y \right) \right)\ge f\left( {{x}^{2}}+{{y}^{2}}+2 \right) \\
\end{aligned} \right.\Rightarrow 4x+4y\ge {{x}^{2}}+{{y}^{2}}+2\Leftrightarrow {{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 6.$
$\Rightarrow \left\{ \begin{aligned}
& {{\left( x-2 \right)}^{2}}\le 6 \\
& {{\left( y-2 \right)}^{2}}\le 6 \\
& x,y\in \mathbb{Z} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& -\sqrt{6}+2\le x\le 2+\sqrt{6} \\
& -\sqrt{6}+2\le y\le 2+\sqrt{6} \\
& x,y\in \mathbb{Z},x>0,y>0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=1,2,3,4 \\
& y=1,2,3,4 \\
\end{aligned} \right..$
Thay $(x;y)$ thảo điều kiện $\left\{ \begin{aligned}
& x=1,2,3,4 \\
& y=1,2,3,4 \\
\end{aligned} \right. $và $ {{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 6 $thì có 15 cặp $ (x;y) $là $ (x;y)\in \left\{ \left( 1;1 \right),\left( 1;2 \right),\left( 1;3 \right),\left( 1;4 \right),\left( 2;1 \right),\left( 2;2 \right),\left( 2;3 \right),\left( 2;4 \right),\left( 3;1 \right),\left( 3;2 \right),\left( 3;3 \right),\left( 3;4 \right),\left( 4;1 \right),\left( 4;2 \right),\left( 4;3 \right) \right\}.$