a,
ĐK: $x\ne y$
Ta có: $x\ge 0, y\ge 0\to x+y\ge 0$
$\dfrac{2}{x^2-y^2}.\sqrt{ \dfrac{3(x+y)^2}{2}}$
$=\dfrac{2}{(x+y)(x-y)}.\sqrt{3}.\dfrac{x+y}{\sqrt2}$
$=\dfrac{\sqrt2.\sqrt3.(x+y)}{(x-y)(x+y)}$
$=\dfrac{\sqrt6}{x-y}$
b,
Ta có: $a>0,5\to 2a-1>0$
$\dfrac{2}{2a-1}.\sqrt{5a^2(1-4a+4a^2)}$
$=\dfrac{2}{2a-1}.\sqrt{5a^2(1-2a)^2}$
$=\dfrac{2}{2a-1}.\sqrt5.|a.(1-2a)|$
$=\dfrac{2}{2a-1}.\sqrt5.a.(2a-1)$
$=2a\sqrt5$
