Question

Resolva as multiplicações e divisões de radicais: (mesmo índice) A) √15 . √8= B) √256 : √16=

Answer 1

Lembrando das seguintes propriedades:

$\large\fbox{ a^b\cdot a^c~\Leftrightarrow~a^{b+c} }\\\\\large\fbox{ (a^m)^n~\Leftrightarrow~a^{m\cdot n} }\\\\\large\fbox{ \sqrt[n]{a^m}~\Leftrightarrow~a^{\frac{m}{n}} } }\\\\\large\fbox{ \sqrt[n]{a^n}~\Leftrightarrow~a^{\frac{n}{n}}~\Leftrightarrow~a^1~\Leftrightarrow~a }\\\\\fbox{ \sqrt[n]{a}\cdot\sqrt[n]{b}~\Leftrightarrow~\sqrt[n]{a\cdot b} }$


A)

$\sqrt{15}\cdot\sqrt{8}~\Leftrightarrow~\sqrt{15\cdot8}~\Leftrightarrow~\sqrt{120}~\Leftrightarrow~\sqrt{4\cdot30}~\Leftrightarrow~\sqrt{2^2\cdot30}~\Leftrightarrow~\sqrt{2^2}\cdot\sqrt{30}\\\\=\fbox{ 2\sqrt{30} }$

B)

$\dfrac{\sqrt{256}}{\sqrt{16}}~\Leftrightarrow~\dfrac{\sqrt{2^8}}{\sqrt{2^4}}~\Leftrightarrow~\dfrac{\sqrt{2^2\cdot2^2\cdot2^2\cdot2^2}}{\sqrt{2^2}\cdot\sqrt{2^2}}~\Leftrightarrow~\dfrac{\sqrt{2^2}\cdot\sqrt{2^2}\cdot\sqrt{2^2}\cdot\sqrt{2^2}}{2\cdot2}\\\\\\\dfrac{2\cdot2\cdot2\cdot2}{4}~\Leftrightarrow~\dfrac{16}{4}=\fbox{ 4 }$