91Ó°ÊÓ

We have given,

$$\begin{gathered} a)\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}} \\ b)\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}} \end{gathered}$$

Here we will use

Quotient rule of the exponent:

$\frac{x^a}{x^b}=x^{a-b}$

Radical rule:

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$

We have,

$\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}$

Applying radical rule,

$\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}=\sqrt{\frac{320m{n}^{-5}}{45m^{-7}{n}^3}}$

Using exponent rule to simplify the fraction inside the radical,

$\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}=\sqrt{\frac{64·5m^{1+7}}{9·5n^{3+5}}}$

role="math" localid="1645705656206" $⇒\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}=\sqrt{\frac{8^2{m}^8}{3^2{n}^8}}$

Using radical rule,

$⇒\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}=\frac{8m^4}{3n^4}$

Hence, simplifying $\frac{\sqrt{320m{n}^{-5}}}{\sqrt{45m^{-7}{n}^3}}$we get,$\frac{8m^4}{3n^4}$.

We have,

$\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}$

Applying radical rule,

$\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}=\sqrt[3]{\frac{16x^4{y}^{-2}}{-54x^{-2}{y}^4}}$

Using exponent rule,

$\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}=\sqrt[3]{\frac{8·2x^{4+2}}{-27·2y^{4+2}}}$$⇒\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}=\sqrt[3]{\frac{8^3{x}^6}{-3^3{y}^6}}$

Using radical rule,

$⇒\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}=-\frac{8x^2}{3y^2}$

Hence, simplifying$\frac{\sqrt[3]{16x^4{y}^{-2}}}{\sqrt[3]{-54x^{-2}{y}^4}}$we will get,$-\frac{8x^2}{3y^2}$.