Giải Hệ Phương Trình: L(x^2 + 1)(y^2 + 1) = 10x + Y)(xy - 1) = 3

Lời giải của Tự Học 365

Giải chi tiết:

Ta có:

\(\begin{array}{l}\;\;\;\;\;\left\{ \begin{array}{l}({x^2} + 1)({y^2} + 1) = 10\\(x + y)(xy - 1) = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^2} + {y^2} + {\left( {xy} \right)^2} + 1 = 10\\\left( {x + y} \right)\left( {xy - 1} \right) = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + y} \right)^2} - 2xy + {\left( {xy} \right)^2} + 1 = 10\\\left( {x + y} \right)\left( {xy - 1} \right) = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {x + y} \right)^2} + {\left( {xy - 1} \right)^2} = 10\\\left( {x + y} \right)\left( {xy - 1} \right) = 3\end{array} \right..\end{array}\)

Đặt: \(\left\{ \begin{array}{l}x + y = u\\xy - 1 = v\end{array} \right.\) thì hệ phương trình trên:

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{u^2} + {v^2} = 10\\uv = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {u + v} \right)^2} - 2uv = 10\\uv = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {u + v} \right)^2} = 16\\uv = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}u + v = 4\\u + v =  - 4\end{array} \right.\\uv = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}u + v = 4\\uv = 3\end{array} \right.\\\left\{ \begin{array}{l}u + v =  - 4\\uv = 3\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}u = 1,\;\;v = 3\\u = 3,\;\;v = 1\\u =  - 1,\;\;v =  - 3\\u =  - 3,\;\;v =  - 1\end{array} \right..\\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + y = 1\\xy - 1 = 3\end{array} \right.\\\left\{ \begin{array}{l}x + y = 3\\xy - 1 = 1\end{array} \right.\\\left\{ \begin{array}{l}x + y =  - 1\\xy - 1 =  - 3\end{array} \right.\\\left\{ \begin{array}{l}x + y =  - 3\\xy - 1 =  - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + y = 1\\xy = 4\end{array} \right.\\\left\{ \begin{array}{l}x + y = 3\\xy = 2\end{array} \right.\\\left\{ \begin{array}{l}x + y =  - 1\\xy =  - 2\end{array} \right.\\\left\{ \begin{array}{l}x + y =  - 3\\xy = 0\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}VN\\x = 2;\;\;y = 1\\x = 1;\;\;y = 2\\x = 1;\;\;y =  - 2\\x =  - 2;\;\;y = 1\\x = 0;\;\;y =  - 3\\x =  - 3;\;\;y = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2;\;\;y = 1\\x = 1;\;\;y = 2\\x = 1;\;\;y =  - 2\\x =  - 2;\;\;y = 1\\x = 0;\;\;y =  - 3\\x =  - 3;\;\;y = 0\end{array} \right..\end{array}\)

Vậy hệ phương trình có tập nghiệm\(S = \left\{ {\left( {1;\;2} \right),\;\left( {2;\;1} \right),\;\left( {1; - 2} \right),\;\left( { - 2;\;1} \right),\;\left( {0; - 3} \right),\;\left( { - 3;\;0} \right)} \right\}.\)

Chọn D.