{({x^5})^{1/3}}{(16{x^3})^{2/3}} {left( {{1 over 4}{x^{4/9}}} right)^{

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Basic of Logarithms

easy

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

${(x/4)^3}$

${(4x)^3}$

$8{x^3}$

એકપણ નહીં

Solution

(d) ${({x^5})^{1/3}}{(16{x^3})^{2/3}}{\left( {{1 \over 4}{x^{4/9}}} \right)^{ – 3/2}}{x^{{5 \over 3} + 3\,.\,{2 \over 3}\, – \,{4 \over 9}.{3 \over 2}}}{2^{{2 \over 3}\,.\,4 + 3}} $

$= {2^{{{17} \over 3}}}{x^3}$.

Standard 11

Mathematics

જો $a = \sqrt {(21)} – \sqrt {(20)} $ અને $b = \sqrt {(18)} – \sqrt {(17),} $ તો

hard

$\root 4 \of {(17 + 12\sqrt 2 )} = $

medium

$\sqrt {(3 + \sqrt 5 )} = . .$ .

medium

જો ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ તો $x = .. . .$

medium

જો $x = {2^{1/3}} – {2^{ – 1/3}},$ તો $2{x^3} + 6x = $

medium

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

Solution

Similar Questions

જો $a = \sqrt {(21)} – \sqrt {(20)} $ અને $b = \sqrt {(18)} – \sqrt {(17),} $ તો

$\root 4 \of {(17 + 12\sqrt 2 )} = $

$\sqrt {(3 + \sqrt 5 )} = . .$ .

જો ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ તો $x = .. . .$

જો $x = {2^{1/3}} – {2^{ – 1/3}},$ તો $2{x^3} + 6x = $

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