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B

- Hướng dẫn giải

Phương pháp giải:

Đặt ẩn phụ. 

Giải chi tiết:

Đặt ẩn phụ \(\left\{ \begin{array}{l}u = 2x + y\\v = 2x - y \ne 0\end{array} \right.\) ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}{u^2} - 5uv + 6{v^2} = 0\,\\u + \frac{1}{v} = 3\,\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {u - 2v} \right)\left( {u - 3v} \right) = 0\,\\u + \frac{1}{v} = 3\,\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}u = 2v\,\\u + \frac{1}{v} = 3\,\end{array} \right.\\\left\{ \begin{array}{l}u = 3v\,\\u + \frac{1}{v} = 3\,\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}u = 2v\,\\2v + \frac{1}{v} = 3\,\end{array} \right.\\\left\{ \begin{array}{l}u = 3v\,\\3v + \frac{1}{v} = 3\,\end{array} \right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}u = 2v\,\\2{v^2} - 3v + 1 = 0\,\end{array} \right.\\\left\{ \begin{array}{l}u = 3v\,\\3{v^2} - 3v + 1 = 0\,\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}u = 2\,\\v = 1\,\end{array} \right.\\\left\{ \begin{array}{l}u = \,1\\v = \frac{1}{2}\,\end{array} \right.\end{array} \right.\end{array}\)

TH1: \(\left\{ \begin{array}{l}u = 2\\v = 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2x + y = 2\\2x - y = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{3}{4}\\y = \frac{1}{2}\end{array} \right.\)

TH2: \(\left\{ \begin{array}{l}u = 1\\v = \frac{1}{2}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}2x + y = 1\\2x - y = \frac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = \frac{3}{8}\\y = \frac{1}{4}\end{array} \right.\)

Vậy hệ phương trình có 2 nghiệm là: \(\left( {\frac{3}{4};\frac{1}{2}} \right),\left( {\frac{3}{8};\frac{1}{4}} \right)\)

Chọn B