CMR với mọi n ∈ z thì a,n*(n+5)-(n-3)*(n+2) chia hết cho 6 b,(n-1)*(n+1)-(n-7)*(n-5) chia hết cho 12 12/09/2021 Bởi Madeline CMR với mọi n ∈ z thì a,n*(n+5)-(n-3)*(n+2) chia hết cho 6 b,(n-1)*(n+1)-(n-7)*(n-5) chia hết cho 12
$a) n.(n+5) – (n-3).(n+2)$ $= n^2 + 5n – (n^2 – 3n + 2n -6)$ $= n^2 + 5n – n^2 + 3n – 2n + 6$ $= (n^2-n^2) + (5n + 3n – 2n) + 6$ $= 6n + 6$ $= 6(n+1) \vdots 6$($đpcm$) $b) (n-1).(n+1)) – (n-7).(n-5)$ $=( n^2 – n + n -1) – (n^2 – 7n – 5n + 35)$ $= n^2 -1 – n^2 + 7n + 5n – 35$ $= -36 + 12n$ $= -12(3-n) \vdots 12$ ($đpcm$) Bình luận
a) Ta có: n.(n + 5) − (n − 3).(n + 2) = n² + 5n − (n² + 2n − 3n − 6) = n² + 5n − n² − 2n + 3n + 6 = 6n + 6 = 6.(n + 1) $\vdots$ 6 (đpcm) b) Ta có: (n – 1).(n + 1) – (n – 7).(n – 5) = n2 + n – n – 1 – n2 + 5n + 7n – 35 = (n2 – n2) + (n – n) + (5n + 7n) – (1 + 35) = 12n – 36 = 12.(n – 3) $\vdots$ 12 (đpcm) Bình luận
$a) n.(n+5) – (n-3).(n+2)$
$= n^2 + 5n – (n^2 – 3n + 2n -6)$
$= n^2 + 5n – n^2 + 3n – 2n + 6$
$= (n^2-n^2) + (5n + 3n – 2n) + 6$
$= 6n + 6$
$= 6(n+1) \vdots 6$($đpcm$)
$b) (n-1).(n+1)) – (n-7).(n-5)$
$=( n^2 – n + n -1) – (n^2 – 7n – 5n + 35)$
$= n^2 -1 – n^2 + 7n + 5n – 35$
$= -36 + 12n$
$= -12(3-n) \vdots 12$ ($đpcm$)
a) Ta có: n.(n + 5) − (n − 3).(n + 2)
= n² + 5n − (n² + 2n − 3n − 6)
= n² + 5n − n² − 2n + 3n + 6
= 6n + 6
= 6.(n + 1) $\vdots$ 6 (đpcm)
b) Ta có: (n – 1).(n + 1) – (n – 7).(n – 5)
= n2 + n – n – 1 – n2 + 5n + 7n – 35
= (n2 – n2) + (n – n) + (5n + 7n) – (1 + 35)
= 12n – 36
= 12.(n – 3) $\vdots$ 12 (đpcm)