ta có : Q=\(\frac{3\sqrt{x}-1}{x-4}=\frac{3\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\);\(R=\frac{2}{\sqrt{x}-2}\)
ĐK:\(x\ge0;x\ne4\)
\(\Rightarrow\frac{Q}{R}=\frac{3\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{2}{\sqrt{x}-2}\)\(=\frac{3\sqrt{x}-1}{2\left(\sqrt{x}+2\right)}=1+\frac{\sqrt{x}-5}{2\left(\sqrt{x}+2\right)}\)
vì 1 \(\in Z\) nên để \(\frac{Q}{R}\in Z\)thì:
\(\frac{\sqrt{x}-5}{2\left(\sqrt{x}+2\right)}\in Z\) \(\Leftrightarrow2\left(\sqrt{x}+2\right)\inƯ\left(\sqrt{x}-5\right)\)
hay \(\sqrt{x}-5⋮2\left(\sqrt{x}+2\right)\Leftrightarrow\sqrt{x}-5⋮2\)
\(\Leftrightarrow\sqrt{x}=2k+5\left(k\in Z\right)\Leftrightarrow x=\left(2k+5\right)^2\)và x\(\ne4\)
vậy x=(2k+5)^2 ; x khác 4 thì Q/R có giá trị nguyên