Ta có: \(\dfrac{9n+21}{3n+2}=\dfrac{9n+6+15}{3n+2}=\dfrac{9n+6}{3n+2}+\dfrac{15}{3n+2}=3+\dfrac{15}{3n+2}\)
Để: \(9n+21⋮3n+2\) thì \(3n+2\inƯ\left(15\right)\)
\(Ư\left(15\right)=\left\{\pm1,\pm3,\pm5,\pm15\right\}\)
Đến đây bn tự lm tiếp nha.
=.= hok tốt!!
