Question

Освободитесь от иррациональности в знаменателе дроби:
1) 10/3√3
2) 18/√13+2

Answer 1

$\displaystyle \tt 1) \: \frac{10}{3\sqrt{3}}=\frac{10}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{10\sqrt{3}}{3\sqrt{3}\sqrt{3}}=\frac{10\sqrt{3}}{3\cdot3}=\frac{10\sqrt{3}}{9}$

$\displaystyle \tt 2) \: \frac{18}{\sqrt{13}+2}=\frac{18}{\sqrt{13}+2}\cdot \frac{\sqrt{13}-2}{\sqrt{13}-2}=\frac{18(\sqrt{13}-2)}{(\sqrt{13}+2)(\sqrt{13}-2)}=\frac{18(\sqrt{13}-2)}{13-4}=\frac{18(\sqrt{13}-2)}{9}=2(\sqrt{13}-2)=2\sqrt{13}-4$

Answer 2

$1)\frac{10}{3\sqrt{3}}=\frac{10*\sqrt{3}}{3\sqrt{3}*\sqrt{3}}=\frac{10\sqrt{3}}{9} \\\\2)\frac{18}{\sqrt{13}+2}=\frac{18*(\sqrt{13}-2)}{(\sqrt{13} +2)(\sqrt{13}-2)}=\frac{18(\sqrt{13}-2)}{(\sqrt{13})^{2}-2^{2}}=\frac{18(\sqrt{13}-2)}{13-4}=\frac{18(\sqrt{13}-2)}{9}=2(\sqrt{13}-2)$