Đáp án đúng: B

Giải chi tiết:\(\begin{array}{l}\,\,\,\,\,\,\,{x^2}{y^2} - 16xy + 99 = 9{x^2} + 36{y^2} + 13x + 26y\\ \Leftrightarrow {x^2}{y^2} + 20xy + 99 = 9{x^2} + 36xy + 36{y^2} + 13x + 26y\\ \Leftrightarrow \left( {{x^2}{y^2} + 20xy + 100} \right) - 1 = {\left( {3x + 2y} \right)^2} + 13\left( {x + 2y} \right)\\ \Leftrightarrow {\left( {xy + 10} \right)^2} - 1 = {\left( {3x + 2y} \right)^2} + 13\left( {x + 2y} \right)\,\,\,\,\left( * \right)\end{array}\)
Đặt \(\left\{ \begin{array}{l}x + 2y = a\,\,\,\,\left( {a > 0} \right)\\xy + 10 = b\,\,\,\left( {b > 10} \right)\end{array} \right.\)
\(\begin{array}{l} \Rightarrow \left( * \right) \Leftrightarrow {b^2} - 1 = 9{a^2} + 13a\\ \Leftrightarrow 9{a^2} + 2.3a.\dfrac{{13}}{6} + \dfrac{{169}}{{36}} - \dfrac{{169}}{{36}} = {b^2} - 1\\ \Leftrightarrow {\left( {3a + \dfrac{{13}}{6}} \right)^2} - {b^2} = \dfrac{{133}}{{36}}\\ \Leftrightarrow {\left( {18a + 13} \right)^2} - 36{b^2} = 133\\ \Leftrightarrow \left( {18a - 6b + 13} \right)\left( {18a + 6b + 13} \right) = 133\,\,\,\left( 1 \right)\end{array}\)
Ta có: \(a,\,\,b > 0\) \( \Rightarrow 18a + 6b + 13 > 18a - 6b + 13 > 0\)
Lại có: \(133 = 133.1 = 19.7\) 
\(\begin{array}{l} \Rightarrow \left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}18a + 6b + 13 = 133\\18a - 6b + 13 = 1\end{array} \right.\\\left\{ \begin{array}{l}18a + 6b + 13 = 19\\18a - 6b + 13 = 7\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}18a + 6b = 120\\18a - 6b =  - 12\end{array} \right.\\\left\{ \begin{array}{l}18a + 6b = 32\\18a - 6b =  - 6\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}12b = 132\\3a - b =  - 2\end{array} \right.\\\left\{ \begin{array}{l}12b = 38\\3a - b =  - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}b = 11\\a = 3\end{array} \right.\\\left\{ \begin{array}{l}b = \dfrac{{19}}{6}\,\,\,\left( {ktm} \right)\\b =  - \dfrac{{25}}{{18}}\,\,\,\left( {ktm} \right)\end{array} \right.\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}x + 2y = 3\\xy + 10 = 11\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\xy = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\y\left( {3 - 2y} \right) = 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\2{y^2} - 3y + 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\\left( {2y - 1} \right)\left( {y - 1} \right) = 0\end{array} \right.\end{array}\)
\( \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\\left[ \begin{array}{l}2y - 1 = 0\\y - 1 = 0\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 3 - 2y\\\left[ \begin{array}{l}y = \dfrac{1}{2}\,\,\left( {ktm} \right)\\y = 1\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 1\,\,\left( {tm} \right)\\y = 1\,\,\left( {tm} \right)\end{array} \right..\)
Vậy phương trình có nghiệm \(\left( {x;\,\,y} \right) = \left( {1;\,\,1} \right).\)
Chọn B.