Có bao nhiêu cặp số nguyên $\left( x;y \right)$ thỏa mãn ${{\log...

Đk: $x-2y>0$ (1)
${{\log }_{2}}\dfrac{1+{{x}^{2}}+{{y}^{2}}}{x-2y}\le {{4}^{x-2y}}-{{2.2}^{{{x}^{2}}+{{y}^{2}}}}+1\Leftrightarrow \text{lo}{{\text{g}}_{2}}(1+{{x}^{2}}+{{y}^{2}})-{{\log }_{2}}(x-2y)\le {{2}^{2x-4y}}-{{2}^{1+{{x}^{2}}+{{y}^{2}}}}+1$
$\Leftrightarrow {{\log }_{2}}\left( 1+{{x}^{2}}+{{y}^{2}} \right)+{{2}^{1+{{x}^{2}}+{{y}^{2}}}}\le {{\log }_{2}}\left( 2x-4y \right)+{{2}^{2x-4y}}$ (*)
Xét hàm số $f\left( t \right)={{\log }_{2}}t+{{2}^{t}},t>0;f'\left( t \right)=\dfrac{1}{t\ln 2}+{{2}^{t}}\ln 2>0,\forall t>0$
Nên hàm số $f\left( t \right)$ đồng biến trên $\left( 0;+\infty \right)$
$\left( * \right)\Leftrightarrow 1+{{x}^{2}}+{{y}^{2}}\le 2x-4y\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}\le 4$ (**).
Có ${{\left( x-1 \right)}^{2}}\le 4\Leftrightarrow -2\le x-1\le 2\Leftrightarrow -1\le x\le 3$. Mà $x\in \mathbb{Z}\Rightarrow x\in \left\{ -1;0;1;2;3 \right\}$
+)Với $x=-1\Rightarrow \left( ** \right)\Leftrightarrow {{\left( y+2 \right)}^{2}}\le 0\Leftrightarrow y=-2(T/m(1))$
+)Với $x=0\Rightarrow \left( ** \right)\Leftrightarrow {{\left( y+2 \right)}^{2}}\le 3,y\in \mathbb{Z}\Rightarrow y\in \left\{ -3;-2;-1 \right\}(T/m(1))$
+)Với $x=1\Rightarrow \left( ** \right)\Leftrightarrow {{\left( y+2 \right)}^{2}}\le 4,y\in \mathbb{Z}\Rightarrow y\in \left\{ -4;-3;-2;-1;0 \right\}(T/m(1))$
+)Với $x=2\Rightarrow \left( ** \right)\Leftrightarrow {{\left( y+2 \right)}^{2}}\le 3,y\in \mathbb{Z}\Rightarrow y\in \left\{ -3;-2;-1 \right\}(T/m(1))$
+)Với $x=3\Rightarrow \left( ** \right)\Leftrightarrow {{\left( y+2 \right)}^{2}}\le 0\Leftrightarrow y=-2(T/m(1))$
Vậy có tất cả 13 cặp số nguyên $\left( x;y \right)$ thỏa đề bài.