Đáp án:

a/ $A=\dfrac{4}{\sqrt{x}+1}$

b/ $x=\dfrac{1}{9}$

Giải thích các bước giải:

a/ `A=(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}).\frac{4}{\sqrt{x}-1}` $(*)$

$\text{ĐKXĐ: $x >0$ và $x \neq 1$}$

$(*) ⇔ A=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{4}{\sqrt{x}-1}$

$=\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{4}{\sqrt{x}-1}$

$=\dfrac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{4}{\sqrt{x}-1}$

$=\dfrac{4}{\sqrt{x}+1}$

b/ $\text{Để $A=3$ thì:}$

$\dfrac{4}{\sqrt{x}+1}=3$

$⇔ 4=3\sqrt{x}+3$

$⇔ 3\sqrt{x}=1$

$⇔ \sqrt{x}=\dfrac{1}{3}$

$⇔ x=\dfrac{1}{9}$ $(TM)$

$\text{Vậy $x=\dfrac{1}{9}$ thì $A=3$}$

a/ $A=(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}).\dfrac{4}{\sqrt{x}-1}$ ($x>0;x\ne 1$)

$=(\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\dfrac{1-\sqrt x}{\sqrt{x}(\sqrt{x}+1)}).\dfrac{4}{\sqrt{x}-1}$

$=\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}.\dfrac{4}{\sqrt{x}-1}$

$=\dfrac{\sqrt{x}(\sqrt{x}-1).4}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}$

$=\dfrac{4}{\sqrt{x}+1}$

b/ $A=3$

$↔\dfrac{4}{\sqrt{x}+1}=3$

$↔\sqrt{x}+1=\dfrac{4}{3}$

$↔\sqrt{x}=\dfrac{1}{3}$

$↔x=\dfrac{1}{9}$