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

Ответ:

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

\[Упростим\ левую\ часть\ тождества:\]

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

\[= \left( \frac{2y + 1^{\backslash y}}{(y + 3)^{2}} - \frac{y - 2^{\backslash y + 3}}{y(y + 3)} \right) \cdot \frac{y^{3} - 9y}{y^{2} + 6\ } =\]

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

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

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

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

\[Что\ и\ требовалось\ доказать.\]