Có bao nhiêu cặp số nguyên $(x, y)$ thỏa mãn $x+y>0,-20 \leq x...

${{\log }_{2}}(x+2y)+{{x}^{2}}+2{{y}^{2}}+3xy-x-y=0\Leftrightarrow {{\log }_{2}}(x+2y)+{{x}^{2}}+2xy+2{{y}^{2}}+xy-x-y=0$
${{\log }_{2}}(x+2y)+x(x+2y)+y(x+2y)+x+y=0\Leftrightarrow {{\log }_{2}}(x+2y)=\left( x+y \right)\left[ 1-\left( x+2y \right) \right].$
Do $\left\{ \begin{aligned}
& x+y>0,-20\le x\le 20 \\
& x+2y>0 \\
& x,y\in Z \\
\end{aligned} \right.\Rightarrow x+2y\ge 1\Rightarrow {{\log }_{2}}\left( x+2y \right)\ge 0.$
Mà $\left\{ \begin{aligned}
& x+y>0 \\
& {{\log }_{2}}\left( x+2y \right)\ge 0 \\
\end{aligned} \right.\Rightarrow 1-\left( x+2y \right)\ge 0\Leftrightarrow x+2y\le 1\Rightarrow x+2y=1.$
Do $\left\{ \begin{aligned}
& 2x+2y>0,x+2y>0 \\
& 2y=1-x \\
& x,y\in Z,-20\le x\le 20 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x>-1,x+2y>0 \\
& x,y\in Z,-20\le x\le 20 \\
& 2y=1-x \\
\end{aligned} \right.\Rightarrow x=1,3,5,7,9,11,13,15,17,19.$
Vậy có 10 cặp $(x, y)$ thỏa mãn yêu cầu bài toán.