${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
$5\sqrt 2 $
$3\sqrt 2 $
$2\sqrt 3 $
$0$
The equation $\sqrt {(x + 1)} - \sqrt {(x - 1)} = \sqrt {(4x - 1)} $, $x \in R$ has
If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $
${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $
${{{{2.3}^{n + 1}} + {{7.3}^{n - 1}}} \over {{3^{n + 2}} - 2{{(1/3)}^{l - n}}}} = $
If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$