`a) A=8n^2-4n+1`
`\qquad A=8n^2+4n-8n-4+5`
`\qquad A=4n(2n+1)-4(2n+1)+5`
`\qquad A=(2n+1)(4n-4)+5`
Để `8n^2-4n+1 \vdots 2n+1`
`=> 5 \vdots 2n+1`
Do `n∈Z=>2n+1∈Ư(5)={+-1;+-5}`
`=> 2n∈{0;-2;4;-6}`
`=> n∈{0;-1;2;-3}`
Vậy `n∈{0;-1;2;-3}` thì A chia hết cho B
`b) A=3n^3+8n^2-15n+6`
`\qquad A=3n^3-n^2+9n^2-3n-12n+4+2`
`\qquad A=n^2(3n-1)+3n(3n-1)-4(3n-1)+2`
`\qquad A=(3n-1)(n^2+3n-4)+2`
Để `3n^3+8n^2-15n+6 \vdots 3n-1`
`=> 2 \vdots 3n-1`
Do `n∈Z=> 3n-1∈Ư(2)={+-1;+-2}`
`=> 3n∈{2;0;3;-1}`
`=> n∈{0;1}`
Vậy `n∈{0;1}` thì A chia hết cho B
`c) A=4n^3-2n^2-6n+5`
`\qquad A=2n^2(2n-1)-(6n-3)+2`
`\qquad A=2n^2(2n-1)-3(2n-1)+2`
`\qquad A=(2n-1)(2n^2-3)+2`
Để `4n^3-2n^2-6n+5 \vdots 2n-1`
`=> 2 \vdots 2n-1`
Do `n∈Z=>2n-1∈Ư(2)={+-1;+-2}`
`=> 2n∈{2;0;3;-1}`
`=> n∈{1;0}`
Vậy `n∈{0;1}` thì A chia hết cho B.
