a/ ĐKXĐ: \(x\ge0,xe1\)
\(A=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
= \(\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{2-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{-1}{\sqrt{x}+1}\)
b/
Thay \(x=\dfrac{4}{9}\) vào A ta được:
\(A=\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}=\dfrac{-3}{5}\)
Vậy khi \(x=\dfrac{4}{9}\) thì \(A=\dfrac{-3}{5}\)
c/ Với \(x\ge0,xe1\)
* Để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\)
\(\Leftrightarrow-2=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\) ( ktmđk)-Loại
Vậy không có giá trị nào của x thỏa mãn \(A=\dfrac{-1}{2}\)
* Để \(A=\dfrac{-1}{4}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{4}\)
\(\Leftrightarrow-4=-\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (tmđk)
Vậy để \(A=\dfrac{-1}{4}\) thì \(x=9\)
