$\root 4 \of {(17 + 12\sqrt 2 )} = $
$\sqrt 2 + 1$
${2^{1/4}}(\sqrt 2 + 1)$
$2\sqrt 2 + 1$
None of these
The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is
If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
The value of $\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $
If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $
${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $