\(a.Q=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)(ĐK:x≥0;x≠1)
\(b.\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(Q=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\frac{\sqrt{3}+2}{\sqrt{3}}=\frac{3+2\sqrt{3}}{3}\)
\(c.Q=\frac{\sqrt{x}+1}{\sqrt{x}-1}=3\Leftrightarrow\sqrt{x}+1=3\sqrt{x}-3\\ \Leftrightarrow2\sqrt{x}=4\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)
\(d.Q=\frac{\sqrt{x}+1}{\sqrt{x}-1}>\frac{1}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{1}{2}>0\\ \Leftrightarrow\frac{2\cdot\left(\sqrt{x}+1\right)}{2\cdot\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}>0\\ \Leftrightarrow\frac{2\sqrt{x}+2-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\\ \Leftrightarrow\frac{\sqrt{x}+3}{\sqrt{x}-1}>0\)
⇔ \(\sqrt{x}-1>0\) (vì \(\sqrt{x}+3>0\forall0< x\ne1\))
\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
\(e.Q=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để Q ∈ Z thì \(\frac{2}{\sqrt{x}-1}\in Z\Leftrightarrow2⋮\sqrt{x}-1hay\sqrt{x}-1\inƯ\left(2\right)\)
Ta có bảng sau
| \(\sqrt{x}-1\) | 1 | -1 | 2 | -2 |
| \(\sqrt{x}\) | 2 | 0 | 3 | -1 |
| \(x\) | 4 | 0 | 9 | (loại) |
Vậy với ......................