ĐK ; \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a, \(Q=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-8\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-7}{\sqrt{x}+1}\)
b. \(Q< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-7}{\sqrt{x}+1}-\frac{1}{2}< 0\Rightarrow\frac{\sqrt{x}-15}{2\left(\sqrt{x}+1\right)}< 0\Rightarrow\sqrt{x}-15< 0\)
\(\Rightarrow0\le x< 225\)và \(x\ne4\)
c. \(Q=\frac{\sqrt{x}-7}{\sqrt{x}+1}=1-\frac{8}{\sqrt{x}+1}\)
Ta thấy \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\frac{-8}{\sqrt{x}+1}\ge-8\Rightarrow1-\frac{8}{\sqrt{x}+1}\ge-7\)
\(\Rightarrow Q\ge-7\)
Vậy \(MinQ=-7\). Dấu bằng xảy ra \(\Rightarrow x=0\)
