If ${a^x} = {b^y} = {(ab)^{xy}},$ then $x + y = $
$0$
$1$
$xy$
None of these
Number of Solution of the equation ${(x)^{x\sqrt x }} = {(x\sqrt x )^x}$ are
The equation $\sqrt {(x + 1)} - \sqrt {(x - 1)} = \sqrt {(4x - 1)} $, $x \in R$ has
The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is
${4 \over {1 + \sqrt 2 - \sqrt 3 }} = $
The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is