Докажите тождество ((2y+1)/(y^2-6y+9)-(y-2)/(y^2+3y)):(y^2+6)/(y^3-9y)=(y-3)/(y+3).

Ответ:

\[1)\ \frac{2y + 1}{y^{2} + 6y + 9} - \frac{y - 2}{y^{2} + 3y} =\]

\[= \frac{2y + 1^{\backslash y}}{(y + 3)^{2}} - \frac{y - 2^{\backslash y + 3}}{y(y + 3)} =\]

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

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

\[= \frac{y^{2} + 6}{y(y + 3)^{2}};\]

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

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

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

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

\[Что\ и\ требовалось\ доказать.\]