Представьте в виде дроби (3x+y)*(y/x-3y/(3x+y))

\[\ \frac{3x + y}{y} \cdot \left( \frac{y^{\backslash 3x + y}}{x} - \frac{3y^{\backslash x}}{3x + y} \right) = \frac{3x + y}{y} \cdot\]

\[\cdot \left( \frac{y(3x + y) - 3y \cdot x}{x(3x + y)} \right) =\]

\[= \frac{3x + y}{y} \cdot \frac{3xy + y^{2} - 3xy}{x(3x + y)} =\]

\[= \frac{(3x + y) \cdot y^{2}}{yx(3x + y)} = \frac{y}{x}\]

\[y = - \frac{4}{x}\]

\[Область\ определения\ функции:\ \ \ \]

\[x \in ( - \infty;\ \ 0) \cup (0;\ + \infty).\]

\[y < 0\ \ \ \ при\ \ \ x > 0.\]

\[\left( \frac{3}{25 - a^{2}} + \frac{1}{a^{2} - 10a + 25} \right) \cdot \frac{(5 - a)^{2}}{2} + \frac{3a}{a + 5} =\]

\[= \left( \frac{3^{\backslash a - 5}}{(5 - a)(5 + a)} + \frac{1^{\backslash a + 5}}{(a - 5)^{2}} \right) \cdot \frac{(5 - a)^{2}}{2} + \frac{3a}{a + 5} =\]

\[= \frac{3 \cdot (5 - a) + 5 + a}{(5 - a)^{2}(5 + a)} \cdot \frac{(5 - a)^{2}}{2} + \frac{3a}{a + 5} =\]

\[= \frac{15 - 3a + 5 + a}{(5 - a)^{2}(5 + a)} \cdot \frac{(5 - a)^{2}}{2} + \frac{3a}{a + 5} =\]

\[= \frac{20 - 2a}{(5 - a)²(5 + a)} \cdot \frac{(5 - a)^{2}}{2} + \frac{3a}{a + 5} =\]

\[= \frac{2 \cdot (10 - a) \cdot (5 - a)^{2}}{2{\cdot (5 - a)}^{2}(5 + a)} + \frac{3a}{a + 5} =\]

\[= \frac{10 - a}{5 + a} + \frac{3a}{a + 5} =\]

\[= \frac{10 - a + 3a}{5 + a} = \frac{10 + 2a}{5 + a} = \frac{2 \cdot (5 + a)}{5 + a} =\]

\[= 2 - не\ зависит\ от\ x.\]

\[\frac{5y}{2^{\backslash 6 + 2y} - \frac{7}{6 + 2y}} = \frac{5y}{\frac{2 \cdot (6 + 2y) - 7}{6 + 2y}} =\]

\[= \frac{5y}{\frac{12 + 4y - 7}{6 + 2y}} = \frac{5y}{\frac{5 + 4y}{6 + 2y}} =\]

\[= \frac{5y(6 + 2y)}{5 + 4y} = \frac{30y + 10y^{2}}{5 + 4y}\]

\[4y + 5 \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 + 2y \neq 0\]

\[4y \neq - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2y \neq - 6\]

\[y \neq - \frac{5}{4} \neq - 1\frac{1}{4}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ y} \neq - 3\]

\[Выражение\ имеет\ смысл\ при\ \ \ y\ \neq - 1,25;\ \ \]

\[y \neq - 3.\]