ĐKXĐ : \(x\ge0\)
Có \(\left|C\right|=\dfrac{1}{3}\)
=> \(\left|\dfrac{-1}{\sqrt{x}+1}\right|=\dfrac{1}{3}\)
<=> \(\dfrac{1}{\left|\sqrt{x}+1\right|}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|\sqrt{x}+1\right|=3\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=3\\\sqrt{x}+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-4\left(\text{loại}\right)\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)
Vậy x = 4 thì |C| = 1/3
\(ĐKXĐ:x\ge0 \\|C|=\dfrac{1}{3}->|\dfrac{-1}{\sqrt x+1}|=\dfrac{1}{3} \\Do\ \dfrac{-1}{\sqrt x+1}<0 \\->\dfrac{-1}{\sqrt x+1}=\dfrac{-1}{3} \\->\sqrt x+1=3<=>x=4(thoả)\)
Ta có: \(\left|C\right|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}C=\dfrac{1}{3}\\C=-\dfrac{1}{3}\end{matrix}\right.\)
Với \(C=\dfrac{1}{3}\) ta có:
\(-\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{3}\) (ĐKXĐ: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+1=-3\)
\(\Leftrightarrow\sqrt{x}=-4\)
Vì \(\sqrt{x}\ge0\forall x\)
Mà \(\sqrt{x}=-4\) (vô lí)
Vậy \(x\in\varnothing\)
Với \(C=-\dfrac{1}{3}\) ta có:
\(-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{3}\) (ĐKXĐ: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\) (TMĐK)
Vậy \(x=4\) thì \(\left|C\right|=\dfrac{1}{3}\)
