Giải thích các bước giải:
a.ĐKXĐ: $x\ge 0, x\ne 1$
Ta có:
$A=\left(\dfrac{\sqrt{x}-1}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{x-1}{\sqrt{x}-1}$
$\to A=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}$
$\to A=\dfrac{\sqrt{x}-1-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\left(\sqrt{x}+1\right)$
$\to A=\dfrac{\sqrt{x}-1-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}$
$\to A=\dfrac{-1-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$
$\to A=-\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$
b.Ta có:
$x=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2$
$\to\sqrt{x}=\sqrt{2}+1$
$\to A=-\dfrac{3+2\sqrt{2}+1}{\left(\sqrt{2}+1-1\right)\left(\sqrt{2}+1+1\right)^2}$
$\to A=-\dfrac{4+2\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+2\right)^2}$
$\to A=-\dfrac{2\left(2+\sqrt{2}\right)}{\sqrt{2}\left(\sqrt{2}+2\right)^2}$
$\to A=-\dfrac{\sqrt{2}}{\sqrt{2}+2}$
$\to A=-\dfrac{1}{1+\sqrt{2}}$
c.Để $A=3$
d.Để $A<0$
$\to -\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}<0$
$\to -\dfrac{1}{\sqrt{x}-1}<0$ vì $x\ge 0$
$\to \sqrt{x}-1>0$
$\to \sqrt{x}>1$
$\to x>1$
