Question

anyone please solve it​

Answer 1

Given :-

$\sf \sqrt[4]{a^3} \times \sqrt[5]{a^2}\times \sqrt{a}\\ \\ \longmapsto\sf where \ \ a \geqslant 0$

To find :-

The simplest form of given expression

• Formula used !!

$\bigstar{\boxed{\sf \sqrt[n]{a^b}= a^\frac{b}{n}}}$

$\bigstar{\boxed{\sf a^b\times a^c =a^{b+c}}}$

$\longmapsto\sf \sqrt[4]{a^3}\times \sqrt[5]{a^2}\times \sqrt{a}\\ \\ \longmapsto\sf Now\ \ by\ applying \ formula \\ \\ \longrightarrow\sf {\blue{\sqrt[n]{a^b}=a^\frac{b}{n}}}\\ \\ \longmapsto\sf a^{3\times \frac{1}{4}} \times a^{2\times \frac{1}{5}}\times a^\frac{1}{2} \\ \\ \longmapsto\sf a^\frac{3}{4} \times a^\frac{2}{5}\times a^\frac{1}{2}\\ \\ \longmapsto\sf According \ to \ formula : \\ \\ \longrightarrow\sf {\purple{a^b\times a^ c= a^{b+c}}}\\ \\ \longmapsto\sf a^{\frac{3}{4}+\frac{2}{5}+\frac{1}{2}}\\ \\ \longmapsto\sf a^{\frac{15+8+10}{20}}\\ \\ \longmapsto\sf a^\frac{33}{20}\\ \\ \longmapsto\sf it's \ also \ written \ as \\ \\ \longmapsto\sf a^{33\times \frac{1}{20}}\\ \\ \longmapsto\sf So, a^{33\times \frac{1}{20}}= \sqrt[20]{a^{33}}$

$\boxed{\sf{\purple{ Answer :- \sqrt[20]{a^{33}}}}}$

◆ So the Answer of your Question is option 4

Answer 2

Answer:

$\sqrt[33]{a ^{20} } \: is \: your \: answer$

Step-by-step explanation:

Given:

$\sqrt[4]{a ^{3} } \times \sqrt[5] {{a} ^{2} } \times \sqrt{a}$

Anything in the form of:

$\sqrt[n]{a} \: can \: be \: written \: as \: a ^{ \frac{1}{n} }$

So rewriting the above terms we get:

$a ^{ \frac{3}{4} } \times a^{ \frac{2}{5} } \times a ^{ \frac{1}{2} }$

A property of indices says us that:

${a}^{m} \times {a}^{n} = a ^{m + n}$

By using the property for the above terms, we get :

${a}^{ (\frac{3}{4 } + \frac{2}{5} + \frac{1}{2} ) }$

$a ^{ \frac{15}{20} + \frac{8}{20} + \frac{10}{20} }$

$= a^{ \frac{23}{20} + \frac{10}{20} }$

$= {a}^{ \frac{33}{20} }$

This can be rewritten as the following :

$\sqrt[33]{a ^{20} }$

The answer for the following question is 4th option