Решите уравнение (5y-6)/(4y^2-9)-(3-3y)/(3+2y)=3/(2y-3).

\[\frac{5y - 6}{4y^{2} - 9} - \frac{3 - 3y^{\backslash 2y - 3}}{3 + 2y} = \frac{3^{\backslash 2y + 3}}{2y - 3}\]

\[\frac{5y - 6 - 6y + 6y^{2} + 9 - 9y - 6y - 9}{(2y - 3)(2y + 3)} = 0\]

\[ОДЗ:\ \ \ y \neq \pm 1,5.\]

\[6y^{2} - 16y - 6 = 0\]

\[D = 64 + 36 = 100\]

\[y_{1} = \frac{8 + 10}{6} = 3;\ \ \ \ \]

\[y_{2} = \frac{8 - 10}{6} = - \frac{2}{6} = - \frac{1}{3}\]

\[Ответ:y = - \frac{1}{3};\ \ \ y = 3.\]