Страница 405 номер 1278, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\frac{a}{a^{2} - 1} + \frac{a^{2} + a - 1}{a^{3} - a^{2} + a - 1} + \frac{a^{2} - a - 1}{a^{3} + a^{2} + a + 1} - \frac{2a^{3}}{a^{4} - 1} =\)
\(= \dfrac{a}{(a - 1)(a + 1)} + \dfrac{a^{2} + a - 1}{a^{2}(a - 1) + (a - 1)} +\)
\(+ \dfrac{a^{2} - a - 1}{a^{2}(a + 1) + (a + 1)} - \dfrac{2a^{3}}{\left( a^{2} - 1 \right)\left( a^{2} + 1 \right)} =\)
\(= \dfrac{a\left( a^{2} + 1 \right) + \left( a^{2} + a - 1 \right)(a + 1) + \left( a^{2} - a - 1 \right)(a - 1) - 2a^{3}}{(a - 1)(a + 1)\left( a^{2} + 1 \right)} =\)
\(= \dfrac{\begin{gathered}a^{3} + a + a^{3} + a^{2} - a + a^{2} + a - 1 + {} \\ {} + a^{3} - a^{2} - a - a^{2} + a + 1 - 2a^{3}\end{gathered}}{\left( a^{2} - 1 \right)\left( a^{2} + 1 \right)} =\)
\(= \dfrac{a^{3} + a}{\left( a^{2} - 1 \right)\left( a^{2} + 1 \right)} = \dfrac{a}{a^{2} - 1};\)

2) \(\frac{1}{a^{2} + 5a + 6} + \frac{2a}{a^{2} + 4a + 3} + \frac{1}{(a + 1)^{2} + a + 1} - \frac{2}{a + 3} =\)
\(= \dfrac{1}{(a + 2)(a + 3)} + \dfrac{2a}{(a + 1)(a + 3)} + \dfrac{1}{(a + 1)(a + 2)} - \dfrac{2}{a + 3} =\)
\(= \dfrac{a + 1 + 2a^{2} + 4a + a + 3 - 2a^{2} - 6a - 4}{(a + 1)(a + 2)(a + 3)} =\)
\(= \dfrac{0}{(a + 1)(a + 2)(a + 3)} = 0.\)