1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
2: Ta có: \(C=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(x-2\sqrt{x}+1\right)\)
\(=\sqrt{x}-1\)
3: Thay \(x=8-2\sqrt{7}\) vào C, ta được:
\(C=\sqrt{7}-1-1=\sqrt{7}-2\)
4: Để C=-3 thì \(\sqrt{x}-1=-\sqrt{3}\)(vô lý)
5: Để \(C>-\dfrac{1}{3}\) thì \(\sqrt{x}-1+\dfrac{1}{3}>0\)
\(\Leftrightarrow\sqrt{x}>\dfrac{2}{3}\)
hay \(x>\dfrac{4}{9}\)
6: Để \(C< 2\sqrt{x}+3\) thì \(\sqrt{x}-1-2\sqrt{x}-3< 0\)
\(\Leftrightarrow-\sqrt{x}< 4\)(luôn đúng)
