આપલે પૈકી root 3 of 9 ,root 4 of {11} ,root 6 | vedclass

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Question

(a) $oot 3 \of 9 ,\,\,oot 4 \of {11} ,\,\,oot 6 \of {17} $

$L.C.M.$ of $3, 4, 6$ is $12$

$	herefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1

Vedclass · Accepted Answer

$\root 3 \of 9 $

Vedclass · Answer

(a) $oot 3 \of 9 ,\,\,oot 4 \of {11} ,\,\,oot 6 \of {17} $

$L.C.M.$ of $3, 4, 6$ is $12$

$	herefore \sqrt[3]{9} = {9^{1/3}} = {({9^4})^{1/12}} = {(6561)^{1/12}}$

$oot 4 \of {11} = {(11)^{1/4}}{({11^3})^{1/12}} = {(1331)^{1/12}}$,

$oot 6 \of {17} = {(17)^{1/6}} = {({17^2})^{1/2}} = {(289)^{1/12}}$

Hence, $oot 3 \of 9 $ is the greatest number.

આપલે પૈકી $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ કઈ સંખ્યા મહતમ છે ?

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