Đáp án:
a) $P=x-\sqrt x+1$
b) $P_{\min}=\dfrac{3}{4}$
Giải thích các bước giải:
a) $P=\dfrac{x^2-\sqrt x}{x+\sqrt x+1}-\dfrac{2x+\sqrt x}{\sqrt x}+\dfrac{2(x-1)}{\sqrt x-1}$
$=\dfrac{\sqrt x(\sqrt x^3-1)}{x+\sqrt x+1}-\dfrac{\sqrt x(2\sqrt x+1)}{\sqrt x}+\dfrac{2(\sqrt x-1)(\sqrt x+1)}{\sqrt x-1}$
$=\sqrt x.(\sqrt x-1)-2\sqrt x-1+2\sqrt x+2$
$=x-\sqrt x+1$
b) $P=x-\sqrt x+1$
$=x-2.\sqrt x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}$
$=\bigg{(}\sqrt x-\dfrac{1}{2}\bigg{)}^2+\dfrac{3}{4}$
Ta có: $\bigg{(}\sqrt x-\dfrac{1}{2}\bigg{)}^2\ge 0$
$⇒\bigg{(}\sqrt x-\dfrac{1}{2}\bigg{)}^2+\dfrac{3}{4}\ge \dfrac{3}{4}$
$⇒P\ge \dfrac{3}{4}$
$⇒P_{\min}=\dfrac{3}{4}$
Dấu "=" xảy ra khi: $\bigg{(}\sqrt x-\dfrac{1}{2}\bigg{)}^2=0$
$⇒x=\dfrac{1}{4}$
Vậy $P_{\min}=\dfrac{3}{4}$ khi $x=\dfrac{1}{4}$.
