Question

Alguien pude ayudarme

Answer 1

Al racionalizar los denominadores de las expresiones, se tiene:

$\displaystyle 1.~ \frac{4}{\sqrt7}.\frac{\sqrt7}{\sqrt7}={\bf \frac{4\sqrt7}{7}}\\2.~\frac{\sqrt5}{\sqrt6}.\frac{\sqrt6}{\sqrt6}={\bf \frac{\sqrt5\sqrt6}{6}}\\3.~\frac{2\sqrt8}{\sqrt{27}}.\frac{\sqrt{27}}{\sqrt{27}}={\bf \frac{2\sqrt8\sqrt{27}}{27}}\\4.~Ver~parte~2\\5.~\frac{2+\sqrt8}{3\sqrt{3}}.\frac{\sqrt{3}}{\sqrt{3}}={\bf \frac{(2+\sqrt8)\sqrt{3}}{9}}$

Continuando:

$\displaystyle 6.~ \frac{2\sqrt n}{\sqrt m}.\frac{\sqrt m}{\sqrt m}={\bf \frac{2\sqrt n\sqrt m}{m}}\\7.~\frac{2\sqrt5}{3\sqrt6}.\frac{\sqrt6}{\sqrt6}={\bf \frac{2\sqrt5\sqrt6}{18}}\\8.~\frac{2}{(3+\sqrt{6})}.\frac{(3-\sqrt{6})}{(3-\sqrt{6})}={\bf \frac{2(3-\sqrt6)}{3}}\\9.~\frac{2+\sqrt n}{\sqrt{2}}.\frac{\sqrt2}{\sqrt{2}}={\bf \frac{(2+\sqrt n)\sqrt2}{2}}\\10.~\frac{5}{(4+\sqrt{7})}.\frac{(4-\sqrt{7})}{(4-\sqrt{7})}={\bf \frac{5(4-\sqrt{7})}{9}}$