91Ó°ÊÓ

Given expression:

$\sqrt{\frac{75r^6{s}^8}{48r{s}^4}}.$

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number, localid="1644885439730" $b≠0$for any integer $nâ‰\yen 2$.

$$\begin{gathered} \sqrt{\frac{75r^6{s}^8}{48r{s}^4}}=\sqrt{\frac{75.r^5.r.s^4.s^4}{48r.s^4}} \\ =\sqrt{\frac{25}{16}.r^5.s^4} \\ =\sqrt{{\left(\frac{5}{4}\right)}^2.{\left(r^2\right)}^2.{\left(s^2\right)}^2.r} \\ =\frac{5}{4}\left|r^2{s}^2\right|\sqrt{r} \end{gathered}$$

Given radical expression:

$\sqrt[3]{\frac{24x^{8}y4}{81x^{2}y}}$.

For simplifying the radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number $b≠0$, for any integer $nâ‰\yen 2$.

$$\begin{gathered} \sqrt[3]{\frac{24x^8{y}^4}{81x^{2}y}}=\sqrt[3]{\frac{24.x^6.x^2.y^3.y}{81x^2.y}} \\ =\sqrt[3]{\frac{8}{27}.x^6.y^3} \\ =\sqrt[3]{{\left(\frac{2}{3}\right)}^3.{\left(x^2\right)}^3.y^3} \\ =\sqrt[3]{{\left(\frac{2}{3}\right)}^3}.\sqrt[3]{{\left(x^2\right)}^3}.\sqrt[3]{{\left(y\right)}^3} \\ =\frac{2}{3}{x}^{2}y \end{gathered}$$

Given expression:

$\sqrt[4]{\frac{32m^9{n}^2}{162m{n}^2}}$.

radical expression following formula is used:

$$\begin{gathered} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}and \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \end{gathered}$$

if $\sqrt[n]{a}and\sqrt[n]{b}$are the real number, $b≠0$for any integer $nâ‰\yen 2$.

$$\begin{gathered} \sqrt[4]{\frac{32m^9{n}^2}{162m{n}^2}}=\sqrt[4]{\frac{32}{162}.m^{9-1}.n^{2-2}} \\ =\sqrt[4]{\frac{16}{81}.m^8} \\ =\sqrt[4]{{\left(\frac{2}{3}\right)}^4}.\sqrt[4]{{\left(m^2\right)}^4} \\ =\frac{2}{3}{m}^2 \end{gathered}$$