Bài 9
a) Ta có
$VT = \dfrac{\sqrt{xy}(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})}{\sqrt{xy}}$
$= (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$
$= x-y = VP$
b) Ta có
$VT = \dfrac{\sqrt{4x^2 - 4x + 1}}{2(2x-1)}$
$= \dfrac{\sqrt{(2x-1)^2}}{2(2x-1)}$
Do $x<\dfrac{1}{2}$ nên $2x - 1<0$. Vậy ta có
$VT = \dfrac{1 - 2x}{2(2x-1)}$
$ = \dfrac{-(2x-1)}{2(2x-1)}$
$= \dfrac{-1}{2} = VP$.
