a/ đkxđ: x >=0; x khác 1
\(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=\dfrac{3x+\sqrt{9x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ Q là số nguyên <=> \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\in Z\)
Có: \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
=> \(\sqrt{x}-1\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}-1=\left\{-2;-1;1;2\right\}\)
\(\Leftrightarrow\sqrt{x}=\left\{0;2;3\right\}\)
\(\Leftrightarrow x=\left\{0;4;9\right\}\) (TM)
Vậy.........
