Giải các hệ phương trình:

a) \[\left\{ \begin{array}{l}2\left( {x + y} \right) + 3\left( {x - y} \right) = 4\\\left( {x + y} \right) + 2\left( {x - y} \right) = 5\end{array} \right.\]                                     

b) \[\left\{ \begin{array}{l}2\left( {x - 2} \right) + 3\left( {1 + y} \right) = - 2\\3\left( {x - 2} \right) - 2\left( {1 + y} \right) = - 3\end{array} \right.\]

a) Điều kiện \[x \ne 0\], \[y \ne 0\]

Đặt \[X = \frac{1}{x}\], \[Y = \frac{1}{y}\]; Ta có hệ

\[\begin{array}{l}\left\{ \begin{array}{l}X - Y = 1\\3{\rm{X + 4Y}} = 5\end{array} \right.\\\left\{ \begin{array}{l}4X - 4Y = 4\\3{\rm{X + 4Y}} = 5\end{array} \right.\\\left\{ \begin{array}{l}4{\rm{X}} - 4Y = 4\\7{\rm{X}} = 9\end{array} \right.\end{array}\]

\[\left\{ \begin{array}{l}X = \frac{9}{7}\\Y = \frac{2}{7}\end{array} \right.\,hay\,\,\left\{ \begin{array}{l}\frac{1}{x} = \frac{9}{7}\\\frac{1}{y} = \frac{2}{7}\end{array} \right..\,\,Suy\,\,ra:\left\{ \begin{array}{l}x = \frac{7}{9}\\y = \frac{7}{2}\end{array} \right.\]

Hệ có nghiệm duy nhất \[\left( {\frac{7}{9};\,\frac{7}{2}} \right)\]

b) Điều kiện \[x \ne 2\], \[y \ne 1\]

Đặt \[X = \frac{1}{{x - 2}}\], \[Y = \frac{1}{{y - 1}}\]. Ta có hệ

\[\begin{array}{l}\left\{ \begin{array}{l}X + Y = 2\\{\rm{2X - 3Y}} = 1\end{array} \right.\\\left\{ \begin{array}{l}3X + 3Y = 6\\{\rm{2X - 3Y}} = 1\end{array} \right.\\\left\{ \begin{array}{l}{\rm{3X + 3}}Y = 6\\{\rm{5X}} = 7\end{array} \right.\\\left\{ \begin{array}{l}X = \frac{7}{5}\\Y = \frac{3}{5}\end{array} \right.\end{array}\]

\[hay\left\{ \begin{array}{l}\frac{1}{{x - 2}} = \frac{7}{5}\\\frac{1}{{y - 1}} = \frac{3}{5}\end{array} \right..\,\,Suyra\left\{ \begin{array}{l}x = \frac{{19}}{7}\\y = \frac{8}{3}\end{array} \right.\]

Hệ có nghiệm duy nhất \[\left( {\frac{{19}}{7};\,\frac{8}{3}} \right)\]