a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A = \(\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(\Leftrightarrow A=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1}{2\left(x-1\right)}-\dfrac{\sqrt{x}-1}{2\left(x-1\right)}-\dfrac{2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(x-1\right)}\)
\(\Leftrightarrow A=\dfrac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{1}{\sqrt{x}+1}\)
b) Khi \(x=\dfrac{4}{9}\) (thảo mãn ĐKXĐ) thì giá trị của A là:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{\dfrac{4}{9}}+1}=-\dfrac{3}{5}\)
Vậy .....
c)
+) Khi \(A=-\dfrac{1}{2}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=1\) (Loại do không thỏa mãn ĐKXĐ)
+) Khi \(A=\dfrac{-1}{4}\) thì ta có:
\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{4}\)
\(\Leftrightarrow x=9\) (thỏa mãn)
Vậy để A = \(-\dfrac{1}{4}\) thì x = 9