Решите номер 7 и 8 , желательно и 9

Решите задачу:

$7a)\; \; \frac{\sqrt7}{\sqrt7-\sqrt5}-\frac{\sqrt5}{\sqrt7+\sqrt5}=\frac{\sqrt7(\sqrt7+\sqrt5)-\sqrt5(\sqrt7-\sqrt5)}{(\sqrt7-\sqrt5)(\sqrt7+\sqrt5)}=\\\\=\frac{7+\sqrt{35}-\sqrt{35}+5}{7-5}= \frac{12}{2}=6 \\\\7b)\; \; \frac{\sqrt2}{\sqrt3-\sqrt2}+\frac{\sqrt3}{\sqrt3+\sqrt2}=\frac{\sqrt2(\sqrt3+\sqrt2)+\sqrt3(\sqrt3-\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=\\\\=\frac{\sqrt6+2+3-\sqrt6}{3-2}=\frac{5}{1}=5\\\\8)\; \; (\sqrt{15}+\sqrt{21})(\sqrt{21}-\sqrt{15})=21-15=6\\\\9)\; \; \frac{(6\sqrt{13})^2}{39}=\frac{6^2\cdot 13}{3\cdot 13}= \frac{36}{3}=12$