a: Ta có: \(Q=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-2}{x-1}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-1}\cdot\dfrac{x-1}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}+2}{\sqrt{x}-2}\)
b: Để Q<0 thì \(\sqrt{x}-2< 0\)
hay x<4
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)
\(a,Q=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x-1}\left(x\ge0;x\ne1\right)\\ Q=\left(\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\\ Q=\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\\ Q=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-2}\)
\(b,Q< 0\Leftrightarrow\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-2}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\left[2\left(\sqrt{x}+1\right)>0\right]\\ \Leftrightarrow x< 4\)
