Nm hc: 2009-2010 Trng : THCS L Qi n Ngi son: Phm Th Thu Hng

Cho mng cc thy c gio v d tit hc hm nay

Mt s phng php chng minh chia ht

Chuyn

nh l v php chia trong tp hp Z. a)nh l: Cho 2 s nguyn a ,b, vi b khc 0, khi c 2 s nguyn q v r duy nht sao cho:a = b.q + r vi| b|>r 0 a: L s b chia, q: L thng. r: l s d. Khi a chia cho b th d c th l 0 ; 1 ; 2; ... | b| -1. .c bit khi r = 0 th a = b.q, ta ni a chia ht cho b hay b l c ca a. K hiu: a b

b) Tnh cht: *Nu a b v b c a c. *Nu tch ab c v ( b,c)=1 a * Nu a b, a c v (b,c)=1 a c. bc

B- Mt s dng ton.1.Dng 1. S sng tnh cht:Trong n s nguyn lin tip c mt v ch mt s chia ht cho n, n 1.2. Dng 2. S dng cc hng ng thc ng nh.

3. Dng 3. S dng php cha c d.4. Dng 4. S dng ng d. 5. Dng 5. Tm ch s tn cng ca mt s. 6. Dng 6. S dng nguyn tc iricl. 7. Dng 7. S dng nh l Fecma

8. Dng 8. S dng phng php qui np.

I.Dng 1. S sng tnh cht: Trong n s nguyn lin tip c mt v ch mt s chia ht cho n, n 1. gii: a) Bi ton 1. Chng minh rng: -Hai s chn lin tip c dng 2n v 2n+2 (vi n Z) a) Tch hai s chn lin tip th chia ht cho 8. nn tch ca chng l: 2n( 2n+2)= ht cho 6. 4n(n+1). b) Tch ba s nguyn lin tip chia -M n v (n+1) l hai s lin tip lin tip nn 120. c) Tch nm s nguyn nguyn chia ht cho c mt s chia ht cho 2 n(n+1) 2 4.n(n+1) 8 -Vy tch ca hai s chn lin tip chia ht cho 8.

(pcm) b) -Gi ba s nguyn lin tip l: (n-1), n,(n+1) (vi n Z) ta c tch ca chng l: A= (n-1)n(n+1). - Trong A c mt bi ca 2, mt bi ca 3, m (2,3)=1 nn: A 2.3 hay A 6. - Vy tch ca ba s nguyn lin tip chia ht cho 6 (pcm)

T bi ton 1b) ta c bi ton cng ni dung sau: Chng minh rng: n3-n chia ht cho 6.

Bi 1 c)-Ta c: 120 = 3.5.8. - Tch ca 5 s nguyn lin tip c dng: A= (n-2) (n-1)n(n+1)(n+2) ( vi n l s nguyn). -Trong 5 s nguyn lin tip c: + Mt bi ca 3 A 3 +Mt bi ca 5 A 5 -Ta cn chng minh A 8.

.

-Tht vy:

+) Nu n l s chn th n v (n+2) l 2 s chn lin tip. +) Nu n l s l th (n+1) v (n-1) l 2 s chn lin tip. Nn: trong A c t nht 2 tha s chn lin tip A 8 ( theo bi ton 1a) - Li c 3, 5 v 8 i mt nguyn t cng nhau nn : T , v suy ra: A 3.5.8 hay A 120 -Vy tch ca 5 s nguyn lin tip chia ht cho 120. (pm)

I.Dng I. S sng tnh chtTrong n s nguyn lin tip c mt v ch mt s chia ht cho n, n 1. -Ta chngn-2)(n-1)n(n+1)(n+2)= n5- 5n3+ 4n. Tr li: ( minh c : (n-2)(n-1)n(n+1)(n+2) hn nhng 120. -Vy ta c bi ton khchia ht cho cng ni dung -Hy th tm hiu xem bi ton trn c th ra di hnh vi bi 1c). thc khc nh th no? Bi ton: Chng minh rng : n5 - 5n3 + 4n chia ht cho 120.

- gii bi ton ny ta dng cc phng phpphn tch a tnc thnh nhn t , khi ta c bi ton quen (Bi 1c).

I.Dng 1. S sng tnh cht: Trong n s nguyn lin tip c mt v ch mt s chia ht cho n, n 1.Gii:ton 2. Chng minh rng: mt tng, trong c mtvihng l * Cch 1. Phn tch n5-n thnh n5-n chia ht cho 30 s n l Bi tch ca 5 s s nguyn.nguyn lin tip, s hng kia c mt tha s bng 5 nh sau: n5-n = n(n4-1) = n(n2-1)(n2+1) = n(n2-1)( n2-4+5) = n(n2-1)(n2-2)+ 5n(n2-1) =(n-2)(n-1)n(n+1)(n+2)+ 5(n-1)n(n+1) ( vi n l s nguyn) -Ta thy: +) (n-2)(n-1)n(n+1)(n+2) l tch ca 5 s nguyn lin tip nn chia ht cho: 2.3.5=30 ( theo bi ton 1c).

+) (n-1)n (n+1) 6 ( Theo bi ton1 b) 5(n-1)n(n+1) 6.5.Suy ra:(n-2)(n-1)n(n+1)(n+2)+ 5(n-1)n(n+1) 30 -Vy : n5-n chia ht cho 30 ( n l s nguyn) (pcm)

Bi 2.Chng minh rng:n5-n chia ht cho 30 vi n l s nguyn. Gii:* Cch 2. Xt hiu gia n5-n v tch ca 5 s nguyn lin tip: (n5 n )- ( n-2)(n-1)n(n+1)(n+2) = (n5- n) ( n5- 5n3+4n)= 5n3- 5n =5( n-1)n(n+1) -Ta thy (n-1)n(n+1) l tch ca 3 s nguyn lin tip,nn: (n-1) n (n+1) 6 5 (n-1)n(n+1) 5.6 (n5-n) (n-2)(n-1)n(n+1)(n+2) 30 -Li c : (n-2)(n-1)n(n+1)(n+2) 30 ( Theo bi 1c) Suy ra : n5 -n chia ht cho 30 (pcm)-T bi ton 2 em c nhn xt g v ch s tn cng ca hiu ( n5 n)? -Tr li: Hiu c tn cng l 0. -T hy nhn xt v ch s tn cng ca n5 v n ? V ta li c bi ton di hnh thc khc nh th no? Bi ton Chng minh rng: n v n5 c ch s tn cng gingnhau ( vi n l s nguyn)

I.Dng 1. S sng tnh cht: Trong n s nguyn lin tipch mt s chia ht cho n, n 1.

c mt v

Bi ton 3. Cho P l s nguyn t ln hn 3, chng minh rng: p2 -1 chia ht cho 24.

Gii: Ta c: p2 -1= (p-1)(p+1)-Do P l s nguyn t ln hn 3 nn P l s l v P khng chia ht cho 3.

-V P l s l nn (p-1) v (p+1) l hai s chn lin tip nn: (p-1)(p+1) chia ht cho 8 ( Theo bi ton 1a). -Mt khc : (p-1)p(p+1) 3 ( Theo bi ton 1b) m P khng chia ht cho 3 (p-1) 3 hoc ( P+1) 3 (p-1)(p+1) 3 . -Li c (3,8)=1, nn t v suy ra (p-1)(p+1) 3.8- Vy (p-1)(p+1) chia ht cho 24 (pcm)

Bi ton 4. Cho m, n l hai s chnh phng l lin tip. Chng minh rng mn-m n +1 chia ht cho 192

Gii: -Hai s chnh phng l lin tip c dng: m = (2k-1)2 v n=(2k+1)2 (vi k -Do : mn-m-n+1 =(m-1)(n-1)= ((2k-1)2-1)((2k+1)2-1)

Z)

=(4k2- 4k)(4k 2 + 4k)=16(k-1)k.k(k+1)

-Ta c:(k-1)k(k+1)(k-1)k.k(k+1)

34

( Theo bi ton 1b) ( Hai tch ca 2 s nguyn lin tip) 12

-Li c (3,4)=1, nn t v suy ra: k2(k-1)(k+1)

16k2(k-1)(k+1) Vy: mn-m-n+1 192 (pcm)

16.12

Bi tp vn dng:Bi1. Chng minh rng: a) Tch ca 4 s nguyn lin tip chia ht cho 24. b) Tch ca 6 s nguyn lin tip chia ht cho 720 c) Tch ca ba s chn lin tip chia ht cho 48Bi 2. Cho a, bl hai s l khng chia ht cho 3. Chng minh rng: a2-b2 chia ht cho 24 Bi 3. Chng minh rng : a) n4+6n3+11n2+6n chia ht cho 24. ( Vi n l s nguyn)

b) n4-4n3-4n2+16n chia ht cho 384 (vi n l s chn, n>4) Bi 4 Cho a, b l cc s nguyn khng chia ht cho 5.Chng minh rng :a5-b5 chia ht cho 5 Bi 5. Chng minh rng tch ca ba s nguyn lin tip trong s gia l lp phng ca mt s t nhin. th chia ht cho 504.

Phn ny thc hin trong 60 pht. Cc phn tip theo s c gi sau.