Giải thích các bước giải:
ĐKXĐ: $x\ge 0;x\ne 1$
a)
$P=\bigg{(}\dfrac{2}{\sqrt x-1}-\dfrac{5}{x+\sqrt x-2}\bigg{)}:\bigg{[}1+\dfrac{3-x}{(\sqrt x-1)(\sqrt x+2)}\bigg{]}\\\,\,\,\,=\bigg{[}\dfrac{2(\sqrt x+2)}{(\sqrt x-1)(\sqrt x+2)}-\dfrac{5}{(\sqrt x-1)(\sqrt x+2)}\bigg{]}:\bigg{[}\dfrac{(\sqrt x-1)(\sqrt x+2)}{(\sqrt x-1)(\sqrt x+2)}+\dfrac{3-x}{(\sqrt x-1)(\sqrt x+2)}\bigg{]}$
$=\dfrac{2\sqrt x-1}{(\sqrt x-1)(\sqrt x+2)}:\bigg{[}\dfrac{x+\sqrt x-2}{(\sqrt x-1)(\sqrt x+2)}+\dfrac{3-x}{(\sqrt x-1)(\sqrt x+2)}\bigg{]}$
$=\dfrac{2\sqrt x-1}{(\sqrt x-1)(\sqrt x+2)}.\dfrac{(\sqrt x-1)(\sqrt x+2)}{\sqrt x+1}$
$=\dfrac{2\sqrt x-1}{\sqrt x+1}$
b) $x=6-2\sqrt 5=5-2\sqrt 5+1=(\sqrt 5-1)^2$
$⇒\sqrt x=\sqrt{(\sqrt 5-1)^2}=|\sqrt 5-1|=\sqrt 5-1$
$P=\dfrac{2\sqrt x-1}{\sqrt x+1}=\dfrac{2\sqrt 5-2-1}{\sqrt 5-1+1}=\dfrac{10-3\sqrt 5}{5}$
Vậy khi $x=6-2\sqrt 5$ thì $P=\dfrac{10-3\sqrt 5}{5}$
c) $P=\dfrac{2\sqrt x-1}{\sqrt x+1}=\dfrac{1}{\sqrt x}$
$⇒2x-\sqrt x=\sqrt x+1$
$⇒2x-2\sqrt x-1=0$
$⇒\left[ \begin{array}{l}\sqrt x=\dfrac{1+\sqrt 3}{2}\\\sqrt x=\dfrac{1-\sqrt 3}{2}\,(L)\end{array} \right.⇒x=\dfrac{2+\sqrt 3}{2}$
Vậy $P=\dfrac{1}{\sqrt x}$ khi $x=\dfrac{2+\sqrt 3}{2}$
d) $P<1-\sqrt x$
$⇒\dfrac{2\sqrt x-1}{\sqrt x+1}<1-\sqrt x$
$⇒\dfrac{2\sqrt x-1}{\sqrt x+1}<\dfrac{(1-\sqrt x)(1+\sqrt x)}{1+\sqrt x}$
$⇒\dfrac{1-x-2\sqrt x+1}{\sqrt x+1}>0$
$⇒\dfrac{-x-2\sqrt x+2}{\sqrt x+1}>0$
$⇒-x-2\sqrt x+2>0$
$⇒-1-\sqrt 3<\sqrt x<-1+\sqrt 3$
$⇒0\le \sqrt x<-1+\sqrt 3$
$⇒0\le x\le 4-2\sqrt 3$
Vậy $P<1-\sqrt x$ khi $0\le x<4-2\sqrt 3$.
