Докажите тождество (a^2/(a+5)-a^3/(a^2+10a+25))∶(a/(a+5)-a^2/(a^2-25))=(5a-a^2)/(a+5).

Ответ:

\[\left( \frac{a^{2}}{a + 5} - \frac{a^{3}}{a^{2} + 10a + 25} \right)\ :\left( \frac{a}{a + 5} - \frac{a^{2}}{a^{2} - 25} \right) =\]

\[= \frac{5a - a²}{a + 5}\]

\[\left( \frac{{a^{2}}^{\backslash a + 5}}{a + 5} - \frac{a^{3}}{(a + 5)^{2}} \right)\ :\left( \frac{a}{a + 5} - \frac{a^{2}}{(a - 5)(a + 5)} \right) =\]

\[= \frac{5a - a²}{a + 5}\]

\[\frac{a^{2}(a + 5) - a³}{(a + 5)²}\ :\frac{a(a - 5) - a²}{(a - 5)(a + 5)} = \frac{5a - a²}{a + 5}\]

\[\frac{(a^{3} + 5a^{2} - a^{3})(a - 5)(a + 5)}{(a + 5)²(a^{2} - 5a - a^{2})} = \frac{5a - a²}{a + 5}\]

\[\frac{5a²(a - 5)}{- 5a(a + 5)} = \frac{5a - a²}{a + 5}\]

\[\frac{a^{2} - 5a}{- (a + 5)} = \frac{5a - a^{2}}{a + 5}\]

\[\frac{5a - a²}{a + 5} = \frac{5a - a²}{a + 5}.\]