The cube root of $9\sqrt 3 + 11\sqrt 2 $ is
$2\sqrt 3 + \sqrt 2 $
$\sqrt 3 + 2\sqrt 2 $
$3\sqrt 3 + \sqrt 2 $
$\sqrt 3 + \sqrt 2 $
The value of $\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $
If ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$then $x =$
${{{{2.3}^{n + 1}} + {{7.3}^{n - 1}}} \over {{3^{n + 2}} - 2{{(1/3)}^{l - n}}}} = $
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $