Question

Решите пожалуйста очень срочно

Answer 1

$1)\; \; \frac{2}{\sqrt1-3}-\frac{7}{\sqrt1-2}=\Big [\; \sqrt1=1\; \Big ]=\frac{2}{-2}-\frac{7}{-1}=-1+7=6$

$2)\; \; \frac{3}{3+\sqrt6}+\frac{2}{2+\sqrt6}=\frac{3(3-\sqrt6)}{(3+\sqrt6)(3-\sqrt6)}+\frac{2(2-\sqrt6)}{(2+\sqrt6)(2-\sqrt6)}=\frac{3(3-\sqrt6)}{9-6}+\frac{2(2-\sqrt6)}{4-6}=\\\\=\frac{3(3-\sqrt6)}{3}+\frac{2(2-\sqrt6)}{-2}=(3-\sqrt6)-(2-\sqrt6)=3-\sqrt6-2+\sqrt6=1$

$3)\; \; \frac{3}{\sqrt7-2}-\frac{2}{\sqrt7+3}-2\sqrt7=\frac{3(\sqrt7+2)}{7-4}-\frac{2(\sqrt7-3)}{7-9}-2\sqrt7=\\\\=\frac{3(\sqrt7+2)}{3}-\frac{2(\sqrt7-3)}{-2}-2\sqrt7=(\sqrt7+2)+(\sqrt7-3)-2\sqrt7=\\\\=2\sqrt7-1-2\sqrt7=-1$

$4)\; \; \frac{1}{3-\sqrt5}+\frac{1}{2-\sqrt5}+\frac{3\sqrt5}{4}=\frac{3+\sqrt5}{9-5}+\frac{2+\sqrt5}{4-5}+\frac{3\sqrt5}{4}=\\\\=\frac{3+\sqrt5}{4}+\frac{2+\sqrt5}{-1}+\frac{3\sqrt5}{4}=\frac{3+\sqrt5-4(2+\sqrt5)+3\sqrt5}{4}=\frac{3+\sqrt5-8-4\sqrt5+3\sqrt5}{4}=\\\\=\frac{3-8}{4}=-\frac{5}{4}=-1,25$