Докажите тождество (2a/(a+3)-4a/(a^2+6a+9)):(a+1)/(a^2-9)-(a^2-9a)/(a+3)=a.

\[\left( \frac{2a}{a + 3} - \frac{4a}{a^{2} + 6a + 9} \right)\ :\frac{a + 1}{a^{2} - 9\ } - \frac{a^{2} - 9a}{a + 3} = a\]

\[Преобразуем\ левую\ часть\ равенства:\]

\[1)\frac{2a^{\backslash a + 3}}{a + 3} - \frac{4a}{(a + 3)^{2}} = \frac{2a^{2} + 6a - 4a}{(a + 3)^{2}} =\]

\[= \frac{2a^{2} + 2a}{(a + 3)^{2}} = \frac{2a(a + 1)}{(a + 3)^{2}}\]

\[2)\ \frac{2a(a + 1)}{(a + 3)^{2}} \cdot \frac{a^{2} - 9}{a + 1} =\]

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

\[3)\ \frac{2a(a - 3)}{a + 3} - \frac{a^{2} - 9a}{a + 3} =\]

\[= \frac{2a^{2} - 6a - a^{2} + 9a}{a + 3} =\]

\[= \frac{a^{2} + 3a}{a + 3} = \frac{a(a + 3)}{a + 3} = a\]

\[a = a\]

\[Что\ и\ требовалось\ доказать.\]