Physics, asked by Gripex, 3 months ago

Simplify :-

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

 \sf  \  a \geqslant 0

Answers

Answered by Anonymous
59

To find :-

• The simplest form of the given expression.

Need to know :-

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

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

Solution :-

Given,

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

 \sf  \  a \geqslant 0

____

Now,

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

 :\implies \sf  a^{3\times \frac{1}{4}} \times a^{2\times \frac{1}{5}}\times a^\frac{1}{2}\\\\

 :\implies \sf a^\frac{3}{4} \times a^\frac{2}{5}\times a^\frac{1}{2}\\\\

:\implies \sf a^{\frac{3}{4}+\frac{2}{5}+\frac{1}{2}}\\\\

 :\implies \sf a^{\frac{15+8+10}{20}}\\\\

:\implies \sf a^\frac{33}{20}\\\\

:\implies \sf a^{33\times \frac{1}{20}}\\\\

:\implies \sf a^{33\times \frac{1}{20}}= \sqrt[20]{a^{33}}\\\\

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

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MяƖиνιѕιвʟє: Wow
Answered by Anonymous
5

 \large \underline{\tt We\:  have\: to \: solve : }

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

 \large \underline{\tt Given :}

 \rm \ a \geqslant 0

 \large \underline{\tt Let's \: Solve :}

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

We know that :

\large \boxed{\tt \sqrt[m]{a} = a^{\frac{1}{m}}}

\rm \implies (a^{3})^{\frac{1}{4}} \times (a^{2})^{\frac{1}{5}} \times a^{\frac{1}{2}}

We know that :

\large \boxed{\tt (a^{m})^{n} = a^{mn}}

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

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

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

We know that :

\large \boxed{\tt a^{m} \times a^{n} = a^{m+n}}

\rm \implies a^{\frac{3}{4} + \frac{2}{5} + \frac{1}{2}}

\rm \implies a^{\frac{15}{20} + \frac{8}{20} + \frac{10}{20}}

\rm \implies a^{\frac{15 + 8 + 10}{20}}

\rm \implies a^{\frac{33}{20}}

 \rm So, \: answer \: is \: a^{\frac{33}{20}}

We can solve it more, by using this formula :

\large \boxed{\tt a^{\frac{1}{m}} = \sqrt[m]{a}}

\rm \implies \sqrt[20]{a^{33}}

 \rm So, \: answer \: is \: \sqrt[20]{a^{33}}

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