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