30 баллов!Срочно надо сдать,помогите4.15

Решите задачу:

$\displaystyle \tt 1). \ \ \frac{5a}{\sqrt{5}}=\frac{5a\cdot \sqrt{5}}{\sqrt{5}\cdot \sqrt{5}}=\frac{5a\cdot \sqrt{5}}{5}=a\sqrt{5};\\\\\\2). \ \ \frac{3+3\sqrt{a}}{\sqrt{a}}=\frac{(3+3\sqrt{a})\cdot \sqrt{a}}{\sqrt{a}\cdot \sqrt{a}}=\frac{3\sqrt{a}+3a}{a}=\frac{3\sqrt{a}}{a}+3;\\\\\\3). \ \ \frac{15c}{\sqrt{3}}=\frac{15c\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}=\frac{15c\cdot \sqrt{3}}{3}=5\sqrt{3}\cdot c;$

$\displaystyle \tt 4). \ \ \frac{5a}{\sqrt{5}-1}=\frac{5a\cdot (\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{5a\cdot (\sqrt{5}+1)}{5-1}=1,25a\cdot(\sqrt{5}+1);\\\\\\5). \ \ \frac{3a}{\sqrt{5}-\sqrt{a}}=\frac{3a\cdot (\sqrt{5}+\sqrt{a})}{(\sqrt{5}-\sqrt{a})(\sqrt{5}+\sqrt{a})}=\frac{3a\cdot (\sqrt{5}+\sqrt{a})}{5-a};\\\\\\6). \ \ \frac{7}{1+\sqrt{3}}=\frac{7\cdot (1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{7\cdot (1-\sqrt{3})}{1-3}=\frac{7\sqrt{3}-7}{2}=3,5\cdot (\sqrt{3}-1);$