Помогите решить 109 номер .

Решите задачу:

$1)\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\\\\=\frac{(\sqrt{5}-\sqrt{3})*(\sqrt{5}-\sqrt{3})+(\sqrt{5}+\sqrt{3})*(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})*(\sqrt{5}-\sqrt{3})}=\\\\=\frac{5-2*\sqrt{5*3}+3+5+2*\sqrt{5*3}+3}{(\sqrt{5})^2-(\sqrt{3})^2}=\\\\=\frac{16}{5-3}=\frac{16}{2}=8$

$2)\frac{11+\sqrt{21}}{11-\sqrt{21}}+\frac{11-\sqrt{21}}{11+\sqrt{21}}=\\\\=\frac{(11+\sqrt{21})*(11+\sqrt{21})+(11-\sqrt{21})*(11-\sqrt{21})}{(11-\sqrt{21})*(11+\sqrt{21})}=\\\\=\frac{121+2*11*\sqrt{21}+21+121-2*11*\sqrt{21}+21}{(11)^2-(\sqrt{21})^2}=\\\\=\frac{242+42}{121-21}=\frac{284}{100}=2,84$

$[tex]3)\frac{1}{11-2\sqrt{30}}-\frac{1}{11+2\sqrt{30}}=\\\\=\frac{1*(11+2\sqrt{30})-1*(11-2\sqrt{30})}{(11-2\sqrt{30})*(11+2\sqrt{30})}\\\\=\frac{11+2\sqrt{30}-11+2\sqrt{30}}{(11)^2-(2\sqrt{30})^2}=\\\\=\frac{4\sqrt{30}}{121-4*30}=\frac{4\sqrt{30}}{121-120}=\frac{4\sqrt{30}}{1}=4\sqrt{30}$

$4)\frac{5}{3+2\sqrt{2}}+\frac{5}{3-2\sqrt{2}}=\\\\=\frac{5*(3-2\sqrt{2})+5*(3+2\sqrt{2})}{(3+2\sqrt{2})*(3-2\sqrt{2})}=\\\\=\frac{15-6\sqrt{2}+15+6\sqrt{2}}{(3)^2-(2\sqrt{2})^2}=\\\\\frac{30}{9-4*2}=\frac{30}{9-8}=\frac{30}{1}=30$