Докажите тождество: $$(\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25}):(\frac{a}{a+5}-\frac{a^2}{a^2-25})=\frac{5a-a^2}{a+5}.$$

Докажем тождество: $$(\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25}):(\frac{a}{a+5}-\frac{a^2}{a^2-25})=\frac{5a-a^2}{a+5}.$$

$$\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25} = \frac{a^2}{a+5} - \frac{a^3}{(a+5)^2} = \frac{a^2(a+5)-a^3}{(a+5)^2} = \frac{a^3+5a^2-a^3}{(a+5)^2} = \frac{5a^2}{(a+5)^2}$$

$$\frac{a}{a+5}-\frac{a^2}{a^2-25} = \frac{a}{a+5} - \frac{a^2}{(a-5)(a+5)} = \frac{a(a-5)-a^2}{(a+5)(a-5)} = \frac{a^2-5a-a^2}{(a+5)(a-5)} = \frac{-5a}{(a+5)(a-5)}$$

$$(\frac{a^2}{a+5}-\frac{a^3}{a^2+10a+25}):(\frac{a}{a+5}-\frac{a^2}{a^2-25}) = \frac{5a^2}{(a+5)^2} : (\frac{-5a}{(a+5)(a-5)}) = \frac{5a^2}{(a+5)^2} \cdot \frac{(a+5)(a-5)}{-5a} = \frac{a(a-5)}{-(a+5)} = \frac{-a(a-5)}{a+5} = \frac{5a-a^2}{a+5}$$

Следовательно, тождество доказано.

Ответ: $$\frac{5a-a^2}{a+5}$$