\[\boxed{\text{1325.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} \frac{2}{y - 1} + \frac{3}{x + 1} = \frac{5}{2} \\ \frac{1}{x - 2} = - \frac{3}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} y = 6 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{2}{6 - 3x - 1} + \frac{3}{x + 1} = \frac{5}{2} \\ \end{matrix} \right.\ \]
\[\frac{2}{5 - 3x} + \frac{3}{x + 1} =\]
\[= \frac{5}{2}\ \ \ | \cdot 2(5 - 3x)(x + 1)\]
\[4(x + 1) + 6(5 - 3x) =\]
\[= 5 \cdot (5 - 3x)(x + 1)\]
\[4x + 4 + 30 - 18x =\]
\[= 5 \cdot \left( 5x - 3x^{2} + 5 - 3x \right)\]
\[34 - 14x = 10x - 15x^{2} + 25\]
\[15x^{2} - 24x + 9 = 0\ \ \ \ |\ :3\]
\[5x^{2} - 8x - 3 = 0\]
\[D_{1} = 16 - 15 = 1\]
\[x_{1} = \frac{4 + 1}{5} = 1;\ \ \ \ \ \ \ \ \ \]
\[\ x_{2} = \frac{4 - 1}{5} = \frac{3}{5} = 0,6.\]
\[y_{1} = 6 - 3 \cdot 1 = 3;\ \ \]
\[\ y_{2} = 6 - 3 \cdot 0,6 = 4,2.\]
\[Ответ:(1;3);\ \ (0,6;4,2).\]
\[\textbf{б)}\ \left\{ \begin{matrix} \frac{1}{y + 1} = \frac{2}{x - 1}\text{\ \ \ \ \ \ \ } \\ \frac{4}{x + 2} + \frac{1}{y - 1} = \frac{1}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} x - 1 = 2y + 2\ \ \ \ \ \ \\ \frac{4}{x + 2} + \frac{1}{y - 1} = \frac{1}{3} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 2y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{4}{2y + 3 + 2} + \frac{1}{y - 1} = \frac{1}{3} \\ \end{matrix} \right.\ \]
\[\frac{4}{2y + 5} + \frac{1}{y - 1} = \frac{1}{3}\text{\ \ }\]
\[\ \ \ | \cdot 3(2y + 5)(y - 1)\]
\[12(y - 1) + 3(2y + 5) =\]
\[= (2y + 5)(y - 1)\]
\[12y - 12 + 6y + 15 =\]
\[= 2y^{2} + 5y - 2y - 5\]
\[18y + 3 = 2y^{2} + 3y - 5\]
\[2y^{2} - 15y - 8 = 0\]
\[D = 225 + 64 = 289 = 17^{2}\]
\[y_{1} = \frac{15 + 17}{4} = 8;\ \ \ \ \ \]
\[y_{2} = \frac{15 - 17}{4} = - \frac{1}{2} = - 0,5.\]
\[x_{1} = 2 \cdot 8 + 3 = 19;\ \ \]
\[\ x_{2} = 2 \cdot ( - 0,5) + 3 = 2.\]
\[Ответ:(19;8);\ \ (2; - 0,5).\]