\(\frac{3x-9}{x-4}\)(\(x\ne4\))=\(\frac{3x-12+21}{x-4}\)=\(\frac{3\left(x-4\right)+21}{x-4}\)=3+\(\frac{21}{x-4}\)
x là số nguyên \(\Leftrightarrow x\varepsilon z\)\(\Leftrightarrow\)x-4\(\varepsilon\)Ư(21)
Ư(21)=\(1,-1,3,-3,7,-7,21,-21\)
Với \(x-4=1\Rightarrow x=5\left(TM\right)\)
\(x-4=-1\Rightarrow x=3\left(TM\right)\)
\(x-4=3\Rightarrow x=7\left(TM\right)\)
\(x-4=-3\Rightarrow x=1\left(TM\right)\)
\(x-4=7\Rightarrow x=11\left(TM\right)\)
\(x-4=-7\Rightarrow x=-3\left(TM\right)\)
\(x-4=21\Rightarrow x=25\left(TM\right)\)
\(x-4=-21\Rightarrow x=-17\left(TM\right)\)
Vậy .......