ĐKXĐ: \(x\ge0,x\ne1\)
a) \(C=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}=\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-2\sqrt{x}+2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-1}{\sqrt{x}+1}\)b) Thay \(x=\frac{4}{9}\) vào C\(\Leftrightarrow C=\frac{-1}{\sqrt{\frac{4}{9}}+1}=\frac{-1}{\frac{2}{3}+1}=\frac{-1}{\frac{5}{3}}=\frac{-3}{5}\)
Vậy C=\(-\frac{3}{5}\) khi x=\(\frac{4}{9}\)
c) \(\left|C\right|=\frac{1}{3}\Leftrightarrow\)\(\left[{}\begin{matrix}C=\frac{1}{3}\\C=\frac{-1}{3}\end{matrix}\right.\)
TH1:\(C=\frac{1}{3}\Leftrightarrow\frac{-1}{\sqrt{x}+1}=\frac{1}{3}\Leftrightarrow-3=\sqrt{x}+1\Leftrightarrow\sqrt{x}=-4\left(ktm\right)\)
TH2:\(C=-\frac{1}{3}\Leftrightarrow\frac{-1}{\sqrt{x}+1}=\frac{-1}{3}\Leftrightarrow3=\sqrt{x}+1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4 thì \(\left|C\right|=\frac{1}{3}\)
