Hệ \(\left\{ \begin{array}{l}{x^2} + {y^2} + x + y = 8(1)\\2xy + x + y = 7(2)\end{array} \right.\) có bao nhiêu nghiệm \((x;y)\) thỏa mãn \(x>0\) và \(y>0\)

\(\left\{ {\begin{array}{*{20}{l}}{{x^2} + {y^2} + x + y = 8}\\{2xy + x + y = 7}\end{array}} \right.\)\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{{\left( {x - y} \right)}^2} = 1}\\{2xy + x + y = 7}\end{array}} \right.\)\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\left[ {\begin{array}{*{20}{l}}{x - y = 1}\\{x - y =  - 1}\end{array}} \right.}\\{2xy + x + y = 7}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{x - y = 1}\\{2xy + x + y = 7}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{x - y =  - 1}\\{2xy + x + y = 7}\end{array}} \right.}\end{array}} \right.\)\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{y = x - 1}\\{2x\left( {x - 1} \right) + 2x - 1 = 7}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{y = x + 1}\\{2x\left( {x + 1} \right) + 2x + 1 = 7}\end{array}} \right.}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{y = x - 1}\\{2{x^2} = 8}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{y = x + 1}\\{2{x^2} + 4x - 6 = 0}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{y = x - 1}\\{x = {\rm{\;}} \pm 2}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{y = x + 1}\\{\left[ {\begin{array}{*{20}{l}}{x = 1}\\{x =  - 3}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = 2;y = 1}\\{x =  - 2;y =  - 3}\\{x = 1;y = 2}\\{x =  - 3;y =  - 2}\end{array}} \right.\)

Vì x>0 và y>0 nên hệ có 2 nghiệm thỏa mãn yêu cầu bài toán.