Question

1.114. Найдите значение выражения: на фото​

Answer 1

$\displaystyle\bf\\1)\\\\ \frac{1}{3\sqrt{2}-4} -\frac{1}{3\sqrt{2}+4 } =\frac{3\sqrt{2} +4-3\sqrt{2}+4 }{(3\sqrt{2}-4)\cdot(3\sqrt{2} +4 )} =\\\\\\=\frac{8}{(3\sqrt{2})^{2}-4^{2} } =\frac{8}{18-16} =\frac{8}{2} =4\\\\2)\\\\ \frac{1}{5+2\sqrt{6}} +\frac{1}{5-2\sqrt{6} } =\frac{5-2\sqrt{6} +5+2\sqrt{6} }{(5+2\sqrt{6})\cdot(5-2\sqrt{6} )} =\\\\\\=\frac{10}{5^{2}- (2\sqrt{6})^{2} } =\frac{10}{25-24} =\frac{10}{1} =10$

$\displaystyle\bf\\3)\\\\ \frac{1}{11-2\sqrt{30}} -\frac{1}{11+2\sqrt{30} } =\frac{11+2\sqrt{30} -11+2\sqrt{30} }{(11-2\sqrt{30})\cdot(11+2\sqrt{30} )} =\\\\\\=\frac{4\sqrt{30} }{11^{2}- (2\sqrt{30})^{2} } =\frac{4\sqrt{30} }{121-120} =\frac{4\sqrt{30} }{1} =4\sqrt{30} \\\\4)\\\\\frac{\sqrt{5}-\sqrt{3} }{\sqrt{5} +\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3} }{\sqrt{5} -\sqrt{3} } =$

$\displaystyle\bf\\=\frac{(\sqrt{5}-\sqrt{3})\cdot(\sqrt{5} -\sqrt{3} )+(\sqrt{5}+\sqrt{3})\cdot(\sqrt{5} +\sqrt{3})}{(\sqrt{5} +\sqrt{3} )\cdot(\sqrt{5}-\sqrt{3}) } =\\\\\\=\frac{5-2\sqrt{15}+3 +5+2\sqrt{15}+3 }{(\sqrt{5} )^{2} -(\sqrt{3} )^{2} } = \frac{16}{5-3} =\frac{16}{2} =8$