Đáp án đúng: D

Phương pháp giải:
Đặt ĐKXĐ để hệ phương trình có nghĩa.
Giải hệ phương trình bằng phương pháp đặt ẩn phụ.
Giải chi tiết:ĐKXĐ: \(x
e y.\)
Ta có:  \(\left( I \right) \Leftrightarrow \left\{ \begin{array}{l}\left( {x - y} \right)\left( {x + y} \right) = 6{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \\{\left( {x + y - 1} \right)^2} - \frac{4}{{{{\left( {x - y} \right)}^2}}} - 3 = 0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \end{array} \right.\)
Đặt  \(\left\{ \begin{array}{l}a = x + y\\b = x - y\end{array} \right.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,\left( {b
e 0} \right)\)
Khi đó:  \(\left( I \right) \Leftrightarrow \left\{ \begin{array}{l}ab = 6{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \\{\left( {a - 1} \right)^2} - \frac{4}{{{b^2}}} - 3 = 0{\kern 1pt} {\kern 1pt} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\frac{1}{b} = \frac{a}{6}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( 1 \right)\\{\left( {a - 1} \right)^2} - \frac{{4{a^2}}}{{36}} = 3{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,\,\left( 2 \right)\end{array} \right.{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \)
 \(\begin{array}{l} \Rightarrow \left( 2 \right) \Leftrightarrow 9{\left( {a - 1} \right)^2} - {a^2} = 27\\ \Leftrightarrow 9{a^2} - 18a + 9 - {a^2} - 27 = 0\\ \Leftrightarrow 8{a^2} - 18a - 18 = 0\\ \Leftrightarrow 4{a^2} - 9a - 9 = 0\\ \Leftrightarrow \left( {a - 3} \right)\left( {4a + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}a - 3 = 0\\4a + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}a = 3\\a = \frac{{ - 3}}{4}\end{array} \right.\end{array}\)
+) Với \(a = 3 \Rightarrow b = \frac{6}{3} = 2\) \( \Rightarrow \left\{ \begin{array}{l}x + y = 3\\x - y = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{5}{2}\,\,\,\left( {tm} \right)\\y = \frac{1}{2}\,\,\,\,\left( {tm} \right)\end{array} \right.\)
+) Với \(a = \frac{{ - 3}}{4} \Rightarrow b = 6:\frac{{ - 3}}{4} =  - 8\) \( \Rightarrow \left\{ \begin{array}{l}x + y = \frac{{ - 3}}{4}\\x - y =  - 8\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{{ - 35}}{8}\,\,\,\left( {tm} \right)\\y = \frac{{29}}{8}\,\,\,\,\left( {tm} \right)\end{array} \right.\)
Vậy hệ đã cho có hai nghiệm \(\left( {x;y} \right) = \left\{ {\left( {\frac{{ - 35}}{8};\frac{{29}}{8}} \right);\,\,\,\left( {\frac{5}{2};\frac{1}{2}} \right)} \right\}.\)
Chọn D.