- Điều kiện xác định: \(x\ne3\) và \(x\ne-3\)
\(B=\dfrac{2}{x-3}+\dfrac{5}{x+3}-\dfrac{12}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{12}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{2x+6+5x-15-12}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{7x-21}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{7\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(B=\dfrac{7}{x+3}\)
Để \(\dfrac{7}{x+3}\in Z\)
\(\Leftrightarrow x+3\inƯ\left(7\right)\)
\(\Leftrightarrow x+3\inƯ\left(\pm1;\pm7\right)\)
- \(x+3=1\Rightarrow x=-2\) (tm)
- \(x+3=-1\Rightarrow x=-4\) (tm)
- \(x+3=7\Rightarrow x=4\) (tm)
- \(x+3=-7\Rightarrow x=-10\) (tm)
Vậy có 4 giá trị của x để \(B\in Z.\)