Giải hệ phương trình:

\[\left\{ \begin{array}{l}4xy + 4\left( {{x^2} + {y^2}} \right) + \frac{3}{{{{\left( {x + y} \right)}^2}}} = \frac{{85}}{3}\\2x + \frac{1}{{x + y}} = \frac{{13}}{3}\end{array} \right.\].

Lời giải:

Đặt \(\left\{ \begin{array}{l}x + y = a \ne 0\\x - y = b\end{array} \right.\) thì \(\left\{ \begin{array}{l}{\left( {x + y} \right)^2} = {a^2}\\{\left( {x - y} \right)^2} = {b^2}\end{array} \right.\) hay \[\left\{ \begin{array}{l}{x^2} + 2xy + {y^2} = {a^2}\\{x^2} - 2xy + {y^2} = {b^2}\end{array} \right.\] 

Suy ra \[\left\{ \begin{array}{l}4xy = {a^2} - {b^2}\\2\left( {{x^2} + {y^2}} \right) = {a^2} + {b^2}\end{array} \right.\]

Từ \(\left\{ \begin{array}{l}x + y = a\\x - y = b\end{array} \right.\) suy ra 2x = a + b.

Hệ phương trình đã cho trở thành:

\[\left\{ \begin{array}{l}{a^2} - {b^2} + 2\left( {{a^2} + {b^2}} \right) + \frac{3}{{{a^2}}} = \frac{{85}}{3}\\a + b + \frac{1}{a} = \frac{{13}}{3}\end{array} \right.\]

\[\left\{ \begin{array}{l}3{a^2} + {b^2} + \frac{3}{{{a^2}}} = \frac{{85}}{3}\\a + b + \frac{1}{a} = \frac{{13}}{3}\end{array} \right.\]

\[\left\{ \begin{array}{l}3\left( {{a^2} + 2 + \frac{1}{{{a^2}}}} \right) + {b^2} = \frac{{85}}{3} + 6\\a + b + \frac{1}{a} = \frac{{13}}{3}\end{array} \right.\]

\[\left\{ \begin{array}{l}3{\left( {a + \frac{1}{a}} \right)^2} + {b^2} = \frac{{103}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\a + \frac{1}{a} = \frac{{13}}{3} - b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\]

Thay (2) vào (1) ta được:

\[3{\left( {\frac{{13}}{3} - b} \right)^2} + {b^2} = \frac{{103}}{3}\]

\(3\left( {\frac{{169}}{9} - \frac{{26}}{3}b + {b^2}} \right) + {b^2} = \frac{{103}}{3}\)

\(169 - 78b + 9{b^2} + 3{b^2} = 103\)

\[12{b^2} - 78b + 66 = 0\]

\(b = 1\) hoặc \(b = \frac{{11}}{2}\)

¬ Với \(b = 1,\) thay vào (2) ta có: \[a + \frac{1}{a} = \frac{{13}}{3} - 1 = \frac{{10}}{3}\]

Suy ra \(3{a^2} - 10a + 3 = 0\)

\(a = 3\) (thỏa mãn) hoặc \(a = \frac{1}{3}\) (thỏa mãn).

Khi đó, ta có:

    ⦁ \(\left\{ \begin{array}{l}x + y = 3\\x - y = 1\end{array} \right.\)  suy ra \(\left\{ \begin{array}{l}x = 2\\y = 1.\end{array} \right.\)

    ⦁ \(\left\{ \begin{array}{l}x + y = \frac{1}{3}\\x - y = 1\end{array} \right.\)  suy ra \(\left\{ \begin{array}{l}x = \frac{2}{3}\\y =  - \frac{1}{3}.\end{array} \right.\)

¬ Với \(b = \frac{{11}}{2},\) thay vào (2) ta có: \[a + \frac{1}{a} = \frac{{13}}{3} - \frac{{11}}{2} =  - \frac{7}{6}.\]

Suy ra \(6{a^2} + 7a + 6 = 0\) (phương trình vô nghiệm).

Vậy hệ phương trình đã cho có nghiệm là \(\left( {x;y} \right) \in \left\{ {\left( {2;\,\,1} \right);\,\,\left( {\frac{2}{3};\,\, - \frac{1}{3}} \right)} \right\}.\)