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Решите задачу:

$1)\; \; \frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3} +\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3} = \frac{(\sqrt5-\sqrt3)^2+(\sqrt5+\sqrt3)^2}{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)} =\\\\=\frac{(5-2\sqrt{15}+3)+(5+2\sqrt{15}+3)}{5-3} =\frac{16}{2} =8\\\\2)\quad \frac{1}{11-2\sqrt{30}} - \frac{1}{11+2\sqrt{30}} = \frac{11+2\sqrt{30}-11+2\sqrt{30}}{(11-2\sqrt{30})(11+2\sqrt{30})} =\\\\= \frac{4\sqrt{30}}{121-4\cdot 30} = \frac{4\sqrt{30}}{121-120} =4\sqrt{30}$

$3)\quad \frac{11+\sqrt{21}}{11-\sqrt{21}} + \frac{11-\sqrt{21}}{11+\sqrt{21}}=\frac{(11+\sqrt{21})^2+(11-\sqrt{21})^2}{(11+\sqrt{21})(11-\sqrt{21})}=\\\\= \frac{(32+22\sqrt{21})+(32-22\sqrt{21})}{11^2-21}=\frac{64}{121-21}=\frac{64}{100} =0,64$

$4)\quad \frac{5}{3+2\sqrt2} + \frac{5}{3-2\sqrt2} = \frac{5(3-2\sqrt2)+5(3+2\sqrt2)}{3^2-4\cdot 2} = \frac{30}{9-8} =30$

$5)\quad (3\sqrt5+\sqrt{15})^2-10\sqrt{27}=9\cdot 5+6\sqrt{5\cdot 15}+15-10\sqrt{3\cdot 9}=\\\\=45+15+6\sqrt{5\cdot 5\cdot 3}-10\cdot 3\sqrt3=60+6\cdot 5\sqrt3-30\sqrt3=\\\\=60+30\sqrt3-30\sqrt3=60$