Đáp án:
a) $A=-\dfrac{1}{\sqrt{x}+1}$
b) $A=\dfrac{1-\sqrt{3}}{2}$
c) Vô nghiệm
Giải thích các bước giải:
a) ĐKXĐ: $x\ge0;x\ne1$
$A=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}$
$=\dfrac{1}{2(\sqrt{x}-1)}-\dfrac{1}{2(\sqrt{x}+1)}-\dfrac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+1}{2(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}-1}{2(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{2\sqrt{x}}{2(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{-2\sqrt{x}+2}{2(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\dfrac{-2(\sqrt{x}-1)}{2(\sqrt{x}-1)(\sqrt{x}+1)}$
$=-\dfrac{1}{\sqrt{x}+1}$
Vậy $A=-\dfrac{1}{\sqrt{x}+1}$
b) Thay $x=3$ ta được:
$A=-\dfrac{1}{\sqrt{3}+1}=\dfrac{1-\sqrt{3}}{2}$
Vậy khi $x=3$ thì $A=\dfrac{1-\sqrt{3}}{2}$
c) $|A|=\dfrac{1}{2}$
$⇔\left[\begin{matrix}-\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\\-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{2}\end{matrix}\right.$
$⇔\left[\begin{matrix}\sqrt{x}+1=-2\\\sqrt{x}+1=2\end{matrix}\right.$
$⇔\left[\begin{matrix}\sqrt{x}=-3 (KTM x\ge0) \\\sqrt{x}=1 (KTM x\ne1)\end{matrix}\right.$
Vậy không có $x$ thỏa mãn
