Question

CMR với mọi n E 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

Answer 1

a, n(n+5) - (n-3)(n+2)

= n² + 5n - (n² + 2n - 3n - 6)

= n² + 5n - n² - 2n + 3n + 6

= 6n + 6

= 6(n + 1) chia hết cho 6 (Đpcm)

b, (n-1)(n+1) - (n-7)(n-5)

= n² + n - n - 1 - (n² - 5n - 7n + 35)

= n² - 1 - n² + 12n - 35

= 12n - 36

= 12(n - 3) chia hết cho 12 (Đpcm)

Answer 2

a)   n(n+5)-(n-3)(n+2)

  =n^2+5n-(n^2+2n-3n+6)

  =n^2+5n-n^2-2n+3n-6

  =6n-6

  =6(n-1) chia het cho 6 voi moi n thuoc z

b)  (n-1)(n+1)-(n-7)(n-5)

  =n^2+n-n-1-(n^2-5n-7n+35)

  =n^2-1-n^2+12n-35

  =12n-36

  =12(n-3) chia het cho 12 voi moi n thuoc z

Answer 3

kobiet

Answer 4

ko hay