a,$ĐKXĐ:x \neq 1$
$P=(\dfrac{\sqrt[]x}{\sqrt[]x-1}+\dfrac{\sqrt[]x}{x-1}):(\dfrac{2}{x}-\dfrac{2-x}{x(\sqrt[]x+1)})$
$=(\dfrac{\sqrt[]x(\sqrt[]x+1)+\sqrt[]x}{x-1}):(\dfrac{2(\sqrt[]x+1)-2+x}{x.(\sqrt[]x+1)})$
$=\dfrac{x+2.\sqrt[]x}{x-1}.\dfrac{x.(\sqrt[]x+1)}{x+2.\sqrt[]x}$
$=\dfrac{x.(\sqrt[]x+1)}{x-1}$
$=\dfrac{x.(\sqrt[]x+1)}{(\sqrt[]x-1)(\sqrt[]x+1)}$
$=\dfrac{x}{\sqrt[]x-1}$
b, $x=\dfrac{2}{2-\sqrt[]3}-2.\sqrt[]3$
$=\dfrac{2-2.\sqrt[]3(2-\sqrt[]3)}{2-\sqrt[]3}$
$=\dfrac{2-4.\sqrt[]3+6}{2-\sqrt[]3}$
$=\dfrac{8-4.\sqrt[]3}{2-\sqrt[]3}$
$=\dfrac{4(2-\sqrt[]3)}{2-\sqrt[]3}$
$=4$
Khi đó $P=\dfrac{4}{\sqrt[]4-1}=\dfrac{4}{2-1}=\dfrac{4}{1}=1$
c, $\sqrt[]P=\sqrt[]{\dfrac{x}{\sqrt[]x-1}}≥0$
Dấu $=$ xảy ra $⇔\dfrac{x}{\sqrt[]x-1}=0$
$⇔x=0$
