ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 95

\[\boxed{\mathbf{95}.}\]

\[1)\ 7\sqrt{28} - \sqrt{80} - 2\sqrt{63} +\]

\[+ 3\sqrt{45} = 7 \cdot 2\sqrt{7} - 4\sqrt{5} -\]

\[- 2 \cdot 3\sqrt{7} + 3 \cdot 3\sqrt{5} =\]

\[= 14\sqrt{7} - 4\sqrt{5} - 6\sqrt{7} +\]

\[+ 9\sqrt{5} = 8\sqrt{7} + 5\sqrt{5}\]

\[2)\ 2\sqrt{\frac{1}{6}} - \sqrt{\frac{2}{3}} + 3\sqrt{\frac{1}{15}} + 4\sqrt{\frac{3}{5}} =\]

\[= \sqrt{\frac{4}{6}} - \sqrt{\frac{2}{3}} + \sqrt{\frac{9}{15}} + 4\sqrt{\frac{3}{5}} =\]

\[= \sqrt{\frac{2}{3}} - \sqrt{\frac{2}{3}} + \sqrt{\frac{3}{5}} + 4\sqrt{\frac{3}{5}} =\]

\[= 5\sqrt{\frac{3}{5}} = \sqrt{\frac{75}{5}} = \sqrt{15}\]

\[3)\ \left( \sqrt{18} - 3\sqrt{2} \right)^{2} = 18 -\]

\[- 6\sqrt{18} \cdot \sqrt{2} + 9 \cdot 2 =\]

\[= 18 - 6 \cdot \sqrt{36} + 18 =\]

\[= 36 - 6 \cdot 6 = 36 - 36 = 0\]

\[4)\ \left( 1 - \sqrt{3} \right)^{2}\left( 1 + \sqrt{3} \right) =\]

\[= \left( 1 - \sqrt{3} \right) \cdot (1 - 3) = 1 -\]

\[- \sqrt{3} - 3 + 3\sqrt{3} = 2\sqrt{3} - 2\]

\[5)\ \frac{3^{\backslash\text{√}5\ + \sqrt{6}}}{\sqrt{5} - \sqrt{6}} - \frac{4^{\backslash\text{√}5 - \sqrt{6}}}{\sqrt{5} + \sqrt{6}} =\]

\[= \frac{}{\left( \sqrt{5} - \sqrt{6} \right)\left( \sqrt{5} + \sqrt{6} \right)} =\]

\[= \frac{3\sqrt{5} + 3\sqrt{6} - 4\sqrt{5} + 4\sqrt{6}}{5 - 6} =\]

\[= \frac{7\sqrt{6} - \sqrt{5}}{- 1} = \sqrt{5} - 7\sqrt{6}\]

\[6)\ \frac{2}{1 + \sqrt{3}} + \frac{3}{3 - \sqrt{5}} =\]

\[= \frac{2 \cdot \left( 1 - \sqrt{3} \right)}{\left( 1 + \sqrt{3} \right)\left( 1 - \sqrt{3} \right)} +\]

\[+ \frac{3 \cdot \left( 3 + \sqrt{5} \right)}{\left( 3 - \sqrt{5} \right)\left( 3 + \sqrt{5} \right)} =\]

\[= \frac{2 - 2\sqrt{3}}{1 - 3} + \frac{9 + 3\sqrt{5}}{9 - 5} =\]

\[= \frac{2 - 2\sqrt{3}}{- 2} + \frac{9 + 3\sqrt{5}}{4} =\]

\[= \left( \sqrt{3} - 1 \right)^{\backslash 4} + \frac{9 + 3\sqrt{5}}{4} =\]

\[= \frac{4\sqrt{3} - 4 + 9 + 3\sqrt{5}}{4} =\]

\[= \frac{5 + 4\sqrt{3} + 3\sqrt{5}}{4} = 1,25 +\]

\[+ \sqrt{3} + 0,75\sqrt{5}\]