Ta có: \(B=\dfrac{12x}{3x+2}=\dfrac{12x+8-8}{3x+2}=\dfrac{4.\left(3x+2\right)}{3x+2}-\dfrac{8}{3x+2}=4-\dfrac{8}{3x+2}\)
Để B thuộc Z thì biểu thức trên phải thuộc Z mà 4 đã thuộc Z nên \(\dfrac{8}{3x+2}\in Z\)
Mà \(\dfrac{8}{3x+2}\in Z\)
thì \(3x+2\inƯ\left(8\right)\)
\(\Leftrightarrow3x+2\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)
\(\Rightarrow x\in\left\{-1;0;-2;2\right\}\)
Vậy \(x\in\left\{-1;0;-2;2\right\}\) thì \(B\in Z\)
\(B=\dfrac{12x}{3x+2}=\dfrac{12x+8-8}{3x+2}=\dfrac{12x+8}{3x+2}-\dfrac{8}{3x+2}=\dfrac{4\left(3x+2\right)}{3x+2}-\dfrac{8}{3x+2}=4-\dfrac{8}{3x+2}\)\(B\in Z\Rightarrow8⋮3x+2\)
\(\Rightarrow3x+2\inƯ\left(8\right)\)
\(Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=1\Rightarrow3x=-1\Rightarrow x=-\dfrac{1}{3}\left(KTM\right)\\3x+2=-1\Rightarrow3x=-3\Rightarrow x=-1\left(TM\right)\\3x+2=2\Rightarrow3x=0\Rightarrow x=0\left(KTM\right)\\3x+2=-2\Rightarrow3x=-4\Rightarrow x=-\dfrac{4}{3}\left(KTM\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=4\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\left(KTM\right)\\3x+2=-4\Rightarrow3x=-6\Rightarrow x=-2\left(TM\right)\\3x+2=8\Rightarrow3x=6\Rightarrow x=2\left(TM\right)\\3x+2=-8\Rightarrow3x=-10\Rightarrow x=-\dfrac{10}{3}\left(KTM\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x+2\\\\\\\end{matrix}\right.\)
