Question

Answer 1

a: Để A chia hết cho B thì \(8n^2-4n+1⋮2n+1\)

=>\(8n^2+4n-8n-4+5⋮2n+1\)

=>\(5⋮2n+1\)

=>\(2n+1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{0;-1;2;-3\right\}\)

b: Để A chia hết cho B thì \(3n^3+8n^2-15n+6⋮3n-1\)

=>\(3n^3-n^2+9n^2-3n-12n+4+2⋮3n-1\)

=>\(2⋮3n-1\)

=>\(3n-1\in\left\{1;-1;2;-2\right\}\)

=>\(3n\in\left\{2;0;3;-1\right\}\)

=>\(n\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)

mà n nguyên

nên \(n\in\left\{0;1\right\}\)

c: Để A chia hết cho B thì \(4n^3-2n^2-6n+5⋮2n-1\)

=>\(4n^3-2n^2-6n+3+2⋮2n-1\)

=>\(2⋮2n-1\)

mà 2n-1 lẻ

nên \(2n-1\in\left\{1;-1\right\}\)

=>\(2n\in\left\{2;0\right\}\)

=>\(n\in\left\{1;0\right\}\)