Решите уравнение: (y+3)/(y-3)=(2y+3)/y.

\[\frac{y + 3}{y - 3} = \frac{2y + 3}{y}\]

\[ОДЗ:\ \ y \neq 0\]

\[\ \ \ \ \ \ \ \ \ \ \ y \neq 3\]

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

\[\frac{y(y + 3) - (y - 3)(2y + 3)}{y(y - 3)} = 0\ \]

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

\[= 0\]

\[\frac{y² + 3y - 2y^{2} + 3y + 9}{y(y - 3)} = 0\]

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

\[- y^{2} + 6y + 9 = 0\ \ \ \ \ \ \ | \cdot ( - 1)\]

\[y^{2} - 6y - 9 = 0\]

\[D = b^{2} - 4ac =\]

\[= 36 - 4 \cdot 1 \cdot ( - 9) = 36 + 36 =\]

\[= 72\]

\[y_{1} = \frac{6 + 6\sqrt{2}}{2} = 3 + 3\sqrt{2}\]

\[y_{2} = \ \frac{6 - 6\sqrt{2}}{2} = 3 - 3\sqrt{2}\]

\[Ответ:\ \ y = 3 + 3\sqrt{2}\ \ \ \ и\ \ \]

\[\ y = 3 - 3\sqrt{2}.\]