If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
$0$
$1$
$n$
$2n$
Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution
$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $
${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$
The cube root of $9\sqrt 3 + 11\sqrt 2 $ is
If ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ and ${b^2} = ac$ then $x + z = $