Question

esto es razonalizacion
1+√7 sobre18-√13
9 sobre√3-√ 2
√8 sobre √8-√3
√5-13 sobre √3-√12
√6-8 sobre √10-√18
con el preoceso por favor es urgente por favor

Answer 1

$\frac{1+\sqrt{7}}{18-\sqrt{13}}$
$=\frac{\left(1+\sqrt{7}\right)\left(18+\sqrt{13}\right)}{\left(18-\sqrt{13}\right)\left(18+\sqrt{13}\right)}=\frac{\left(1+\sqrt{7}\right)\left(18+\sqrt{13}\right)}{311}$

$\frac{9}{\sqrt{3}-\sqrt{2}}$
$=\frac{9\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=\frac{9\left(\sqrt{3}+\sqrt{2}\right)}{1}=9\left(\sqrt{2}+\sqrt{3}\right)$

$\frac{\sqrt{8}}{\sqrt{8}-\sqrt{3}}$
$=\frac{\sqrt{8}}{2\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}}{2\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(2\sqrt{2}+\sqrt{3}\right)}{5}$

$\frac{\sqrt{5}-13}{\sqrt{3}-\sqrt{12}}$
$=\frac{13-\sqrt{5}}{\sqrt{12}-\sqrt{3}}=\frac{13-\sqrt{5}}{2\sqrt{3}-\sqrt{3}}=\frac{13-\sqrt{5}}{\sqrt{3}}=\frac{\sqrt{3}\left(13-\sqrt{5}\right)}{3}$

$\frac{\sqrt{6}-8}{\sqrt{10}-\sqrt{18}}$
$=\frac{8-\sqrt{6}}{\sqrt{18}-\sqrt{10}}=\frac{8-\sqrt{6}}{\sqrt{2}\cdot \:3-\sqrt{10}}=\frac{4\sqrt{10}+\sqrt{2}\cdot \:12-3\sqrt{3}-\sqrt{15}}{4}$