Докажите тождество (a/(a^2-25)-(a-8)/(a^2-10a+25))∶(a-20)/(a-5)^2=-a/(a+5).

Ответ:

\[\left( \frac{a}{a^{2} - 25} - \frac{a - 8}{a^{2} - 10a + 25} \right)\ :\frac{a - 20}{(a - 5)²} = - \frac{2}{a + 5}\ \]

\[\left( \frac{a}{(a - 5)(a + 5)} - \frac{a - 8}{(a - 5)^{2}} \right) \cdot \frac{(a - 5)²}{a - 20} = - \frac{2}{a + 5}\]

\[\frac{\left( a(a - 5) - (a - 8)(a + 5) \right)(a - 5)²}{(a - 5)²(a + 5)(a - 20)} = - \frac{2}{a + 5}\]

\[\frac{a² - 5a - a^{2} - 5a + 8a + 40}{(a + 5)(a - 20)} = - \frac{2}{a + 5}\]

\[\frac{- 2a + 40}{(a + 5)(a - 20)} = - \frac{2}{a + 5}\]

\[\frac{- 2 \cdot (a - 20)}{(a + 5)(a - 20)} = - \frac{2}{a + 5}\]

\[- \frac{2}{a + 5} = - \frac{2}{a + 5}.\]

\[Что\ и\ требовалось\ доказать.\]