Решите уравнение 10y/(9y^2-4)+(y-5)/(3y+2)=(y-3)/(2-3y).

\[\frac{10y}{9y^{2} - 4} + \frac{y - 5}{3y + 2} = \frac{y - 3}{2 - 3y}\]

\[\frac{10y}{(3y - 2)(3y + 2)} + \frac{y - 5^{\backslash 3y - 2}}{3y + 2} + \frac{y - 3^{\backslash 3y + 2}}{3y - 2} = 0\]

\[ОДЗ:\ \ y \neq \pm \frac{2}{3}.\]

\[10y + 3y^{2} - 15y - 2y + 10 + 3y^{2} - 9y + 2y - 6 = 0\]

\[6y^{2} - 14y + 4 = 0\ \ \ \ \ |\ :2\]

\[3y^{2} - 7y + 2 = 0\]

\[D = 49 - 24 = 25\]

\[y_{1} = \frac{7 + 5}{6} = 2;\ \ \ \ \]

\[y_{2} = \frac{7 - 5}{6} = \frac{2}{6} = \frac{1}{3}.\]

\[Ответ:y = 2;\ \ y = \frac{1}{3}.\]