ĐK: \(x\ge0\),\(x\ne1\)
a) \(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}-\dfrac{\sqrt{x}}{x-1}=\dfrac{2\sqrt{x}+2}{\left(2\sqrt{x}-2\right)\left(2\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}-2}{\left(2\sqrt{x}-2\right)\left(2\sqrt{x}+2\right)}-\dfrac{4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}+2}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}-2}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}+2-2\sqrt{x}+2-4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{4-4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-4\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\)
b) Ta có x=\(\dfrac{4}{9}\) thỏa mãn ĐK
Thay x=\(\dfrac{4}{9}\) vào A thì A=\(\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\left(\dfrac{2}{3}+1\right)=-1:\dfrac{5}{3}=-\dfrac{3}{5}=-0,6\)
Vậy khi x=\(\dfrac{4}{9}\) thì A=-0,6
c)\(\left|D\right|=\left|\dfrac{-1}{\sqrt{x}+1}\right|=\dfrac{1}{\sqrt{x}+1}\)(vì \(\sqrt{x}+1>0\))
Ta có \(\left|D\right|=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+1=3\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)Vậy x=4 thì \(\left|D\right|=\dfrac{1}{3}\)
