Đáp án: b.$x>2$
Giải thích các bước giải:
a.Ta có :
$P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}$
$\to P=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}$
$\to P=\left(\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}$
$\to P=\dfrac{\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)}{x}$
$\to P=\dfrac{x-1}{x}$
b.Để $P>\dfrac12$
$\to \dfrac{x-1}{x}>\dfrac12$
$\to 1-\dfrac{1}{x}>\dfrac12$
$\to \dfrac12>\dfrac1x$
$\to x>2$ vì $x>0$
