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

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

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

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

\[= \left( \frac{a^{\backslash a + 1}}{(a - 1)^{2}} - \frac{a + 4^{\backslash a - 1}}{(a - 1)(a + 1)} \right) \cdot \frac{a^{2} - 1}{a - 2} =\]

\[= \frac{a^{2} + a - a^{2} - 4a + a + 4}{\left( a^{2} - 1 \right)(a - 1)} \cdot \frac{a^{2} - 1}{a - 2} =\]

\[= \frac{(4 - 2a)\left( a^{2} - 1 \right)}{\left( a^{2} - 1 \right)(a - 1)(a - 2)} =\]

\[= \frac{- 2(a - 2)}{(a - 1)(a - 2)} = - \frac{2}{a - 1} = \frac{2}{1 - a}\]

\[\frac{2}{1 - a} = \frac{2}{1 - a}\]

\[Что\ и\ требовалось\ доказать.\]