ĐKXĐ: \(x\ge0;x\ne1\)
\(C=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(C=\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)-2\sqrt{x}}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(C=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
C=\(\dfrac{2-2\sqrt{x}}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(C=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(C=-\dfrac{1}{\sqrt{x}+1}\)
b) Thay \(x=\dfrac{4}{9}\) vào C
=> C=\(\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}\)
C=\(\dfrac{-1}{\dfrac{2}{3}+1}\)
C=\(\dfrac{-1}{\dfrac{5}{3}}\)
C=\(-\dfrac{3}{5}\)
C) Ta có |C|=1/3
<=> C=1/3 hoặc C=-1/3
TH1: C=1/3
\(\Rightarrow-\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{3}\\
\Leftrightarrow\sqrt{x}+1=-3\\
\Leftrightarrow\sqrt{x}=-4\left(voly\right)\)
TH2: C=-1/3
\(\Rightarrow-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{3}\\
\Leftrightarrow\sqrt{x}+1=3\\
\Leftrightarrow\sqrt{x}=2\\
\Leftrightarrow x=4\left(nhận\right)\)
Vậy s={4}