ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1279

\[\boxed{\mathbf{1279}\mathbf{.}}\]

\[1)\ \frac{1}{4 + 4\sqrt{a}} - \frac{1}{2 - 2a} + \frac{1}{4 - 4\sqrt{a}} =\]

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

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

\[= \frac{1 - \sqrt{a} - 2 + 1 + \sqrt{a}}{4(1 - a)} =\]

\[= \frac{0}{4(1 - a)} = 0\]

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

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

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

\[= \frac{\sqrt{2} + 1\ }{\sqrt{2} + 2\ } = \frac{\left( \sqrt{2} + 1 \right)\left( 2 - \sqrt{2} \right)}{\left( 2 + \sqrt{2} \right)\left( 2 - \sqrt{2} \right)} =\]

\[= \frac{2\sqrt{2} - 2 + 2 - \sqrt{2}}{4 - 2} = \frac{\sqrt{2}}{2}\]