Question

Answer 1

a) Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)

\(=n^3+2n^2+3n^2+6n-n-2-n^3+2\)

\(=5n^2+5n⋮5\)

b) Ta có: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-\left(6n^2-3n+10n-5\right)\)

\(=6n^2+31n+5-6n^2-7n+5\)

\(=24n+10⋮2\)

Answer 2

c.

$n(n+5)-(n-3)(n+2)=n^2+5n-(n^2-n-6)$
$=6n+6=6(n+1)\vdots 6$

(đpcm)

d.

$(n-1)(n+1)-(n-7)(n-5)=(n^2-1)-(n^2-12n+35)$

$=12n-36=12(n-3)\vdots 12$ 

(đpcm)

e.

$(n-1)(n+6)-(n+1)(n-6)$

$=(n^2+5n-6)-(n^2-5n-6)=10n\vdots 10$ (đpcm)