Question

Racionalizando os denominadores das frações abaixo, teremos, respectivamente:

1_a com a>0

p p com p>0


√a/a e √p
√a e √p
√1 e p
√1/a e p


Me ajuda fazendo favor?

Answer 1

$\large\boxed{\begin{array}{l}\rm Racionalizando~os~denominadores~\\\rm das~frac_{\!\!,}\tilde oes~abaixo,teremos,respectivamente\\\rm\dfrac{1}{\sqrt{a}},com~a>0\\\\\rm\dfrac{p}{\sqrt{p}},com~p>0\\\rm a)~\dfrac{\sqrt{a}}{a}~e~\sqrt{p}\\\\\rm b)~\sqrt{a}~e~\sqrt{p}\\\\\rm c)~\sqrt{1}~e~p\\\\\rm d)~\dfrac{\sqrt{1}}{a}~e~p\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm soluc_{\!\!,}\tilde ao:}\\\\\tt a)~\sf\dfrac{1}{\sqrt{a}}\\\underline{\rm multiplicando~e~dividindo~por~\sqrt{a}~temos:}\\\\\sf\dfrac{1}{\sqrt{a}}\cdot\dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{\sqrt{a}}{\sqrt{a^{\backslash\!\!\!2}}}=\dfrac{\sqrt{a}}{a}\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf\dfrac{p}{\sqrt{p}}\\\underline{\rm multiplicando~e~dividindo~por~\sqrt{p}~temos:}\\\\\sf\dfrac{p}{\sqrt{p}}\cdot\dfrac{\sqrt{p}}{\sqrt{p}}=\dfrac{p\sqrt{p}}{\sqrt{p^{\backslash\!\!\!2}}}=\dfrac{\backslash\!\!\!p\sqrt{p}}{\backslash\!\!\!p}=\sqrt{p}\end{array}}$

$\Huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~a}}}}$