${{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 $
$0$
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
If ${x^{x\root 3 \of x }} = {(x\,.\,\root 3 \of x )^x},$ then $x =$
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
If ${a^x} = {b^y} = {(ab)^{xy}},$ then $x + y = $
${a^{m{{\log }_a}n}} = $