Question

Racionalização de denominadores: Qual o resultado de:

a) ab/³√a²=
b) 2x/$\sqrt[6]{x^{5} }$=
c) 2/√7 + √5=

Answer 1

$\boxed{\begin{array}{l}\tt a)~\sf\dfrac{ab}{\sqrt[\sf3]{\sf a^2}}\cdot\dfrac{\sqrt[\sf3]{\sf a}}{\sqrt[\sf3]{\sf a}}=\dfrac{ab\sqrt[\sf3]{\sf a}}{\sqrt[\sf3]{\sf a^3}}\\\sf =\dfrac{\diagup\!\!\!\!ab\sqrt[\sf3]{\sf a}}{\diagup\!\!\!a}=b\sqrt[\sf3]{\sf a}\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf\dfrac{2x}{\sqrt[\sf6]{\sf x^5}}\cdot\dfrac{\sqrt[\sf6]{\sf x}}{\sqrt[\sf6]{\sf x}}=\dfrac{2x\sqrt[\sf6]{\sf x}}{\sqrt[\sf6]{\sf x^6}}\\\\\sf\dfrac{2\diagdown\!\!\!x\sqrt[\sf6]{\sf x}}{\diagdown\!\!\!x}=2\sqrt[\sf6]{\sf x}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{2}{\sqrt{7}+\sqrt{5}}\cdot\dfrac{(\sqrt{7}-\sqrt{5})}{(\sqrt{7}-\sqrt{5})}=\dfrac{2\cdot(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2}\\\\\sf\dfrac{2\cdot(\sqrt{7}+\sqrt{5}}{7-5}=\dfrac{\diagdown\!\!\!\!2\cdot(\sqrt{7}+\sqrt{5})}{\diagdown\!\!\!\!\!2}=\sqrt{7}+\sqrt{5}\end{array}}$