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