Решите уравнение: (y+4)/(y+2)=(2y-1)/y.

\[\frac{y + 4}{y + 2} = \frac{2y - 1}{y}\]

\[ОДЗ:\ \ y \neq 0\]

\[y + 2 \neq 0 \Rightarrow y \neq - 2\]

\[\frac{y + 4}{y + 2} - \frac{2y - 1}{y} = 0\]

\[\frac{y(y + 4) - (y + 2)(2y - 1)}{y(y + 2)} = 0\]

\[y^{2} + 4y - \left( 2y^{2} - y + 4y - 2 \right) =\]

\[= 0\]

\[y^{2} + 4y - 2y^{2} + y - 4y + 2 = 0\]

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

\[y² - y - 2 = 0\]

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

\[= 1 - 4 \cdot 1 \cdot ( - 2) = 9\]

\[y_{1} = \frac{1 + 3}{2} = \frac{4}{2} = 2\]

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

\[Ответ:y = 2\ \ и\ \ y = - 1.\]