[tex]T= \frac{\sqrt{x}-2}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}[/tex]
Dễ thấy [tex]0< \frac{3}{\sqrt{x}+1}\leq \frac{3}{0+1}=3\Rightarrow \frac{3}{\sqrt{x}+1}\in \left \{ 1;2;3 \right \}[/tex]
[tex]\Rightarrow \sqrt{x}+1\in \left \{ 3;\frac{3}{2};1 \right \}\Rightarrow \sqrt{x}\in \left \{ 2;\frac{1}{2};0 \right \}\Rightarrow x\in \left \{ 4;\frac{1}{4};0 \right \}[/tex]