hàm phần nguyên và ứng dụng.pdf

7/26/2019 Hm phn nguyn v ng dng.pdf

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BGIO DCV OTO IHCTHI NGUYN

TRNGIHCKHOA HC

NGUYN TH HNG HNH

HM PHN NGUYN

V NG DNG

Chuyn ngnh: Phng php Ton s cp

M s: 60.46.40

LUNVNTHCSTON HC

Ngi hng dn khoa hc:

PGS. TS. TDuy Phng

THI NGUYN - 2010

S ha bi Trung tm Hc liu - i hc Thi Nguyn http://www.lrc-tnu.edu.vn


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MC LC

Trang

Cc k hiu ........................................................................................................2

Li ni u ....................................................................................................3-4

Chng 1 Cc kin thc c bn v hm phn nguyn ...............................5

1 Khi nim v phn nguyn .........................................................................5

2 Cc tnh cht c bn ca phn nguyn .......................................................6

3 Hm phn nguyn v th ca hm phn nguyn ................................. 11

Chng 2 Phn nguyn trong ton s hc v i s .................................16

1 Phn nguyn trong cc bi ton s hc................................................... 16

2 Tnh gi tr ca mt s hoc mt biu thccha phn nguyn ................27

3 Chng minh cc h thc cha phn nguyn ..............................................31

4 Phng trnh v h phng trnh cha phn nguyn ...............................32

Chng 3 Phn nguyn trong ton gii tch ..............................................49

1 Mt s tnh cht gii tch ca dy cha phn nguyn ..............................492 Tnh tng hu hn ca dy cha phn nguyn .........................................53

3 Tnh gii hn ca dy cha phn d ................................................56

4 Hm s chaphn nguyn ...........................................................62

5 Chui s cha phn nguyn .............................................................67

Kt lun.........................................................................................................77

Ti liu tham kho........................................................................................78



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2

CC K HIU

Trong cun lun vn ny ta s dng cc k hiu sau:

Tp cc s thc c k hiu l .

Tp cc s thc khng m c k hiu l .

Tp cc s hu t c k hiu l .

Tp cc s nguyn c k hiu l {..., -2, -1, 0,1, 2,...} .

Tp cc s t nhin c k hiu l {1, 2, 3,...} .

Tp cc s nguyn dng c k hiu l hoc .



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LI NI U

Do tnh c o ca hm phn nguyn, th d, hm phn nguyn va

n gin (l hm hng tng khc) li va phc tp (gin on ti cc im

nguyn nn kh p dng cc cng c ca gii tch), nhiu bi ton hay v

phn nguyn c s dng lm thi hc sinh gii cc cp, trong c rt

nhiu cc thi hc sinh gii quc gia v Olympic quc t. Mt khc, hm

phn nguyn c nhng ng dng quan trng khng ch trong ton hc ph

thng, m cn trong nhiu vn ca ton ng dng v cng ngh thng tin

(lm trn s, tnh gn ng,...). Phn nguyn cng th hin s kt ni giatnh lin tc v tnh ri rc, gia ton gii tch v ton ri rc nn kh th v.

L thuyt v bi tp v phn nguyn ri rc c trong cc sch v cc

tp ch, thm ch l nhng chuyn trong mt s sch v s hc (xem[3],

[5], [8]). Tuy nhin, hnh nh cha c mt cun sch no vit phong ph

v tng hp v phn nguyn. chnh l l do tc gi chn ti ny lm

lun vn cao hc.

Lun vn Hm phn nguyn v ng dng c mc ch trnh by cc

kin thc c bn ca hm phn nguyn v ng dng ca n trong gii ton s

cp, c th l trong s hc, i s v gii tch (ton chia ht, gii phng

trnh, tnh cht ca dy, tnh gii hn, tnh tng ca dy, chui,...cha phn

nguyn). ng thi lun vn cng trnh by mi quan h mt thit ca phn

nguyn vi cc dng ton khc (dy truy hi, nh thcNewton, h m,...).

c bit lun vn tp hp mt khi lng ln cc bi ton thi v ch quc

gia v quc tminh ha cho l thuytvphnnguyn.

Lun vn gmbachng.

Chng 1 trnh by cc nh ngha v tnh cht c bn ca hm phn

nguyn v th ca hm phn nguyn.



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Chng 2 trnh by mt s dng ton cha phn nguyn trong s hc

v i s (ton chia ht; tnh ton v chng minh cc h thc cha phn

nguyn; gii phng trnh v h phng trnh cha phn nguyn;...).Chng 3 trnh by mt s dng ton cha phn nguyn trong gii tch

(cc tnh cht nh tnh b chn, tnh tun hon ca dy s; tm s hng v tnh

gii hn ca dy s, tnh tng hu hn ca dy s, tnh tng ca chuicha

phn nguyn, ...).

Nhiu v d v bi ton tp hp trong lun vn c a vo bn tho

cun sch ca tc gi lun vn vit chung vi Thy hng dn v Thc s

Nguyn Th Bnh Minh. V hn ch s trang lun vn, trong mi chng,chng ti c gng trnh by cc vn l thuyt lm c s phn loi v

tng kt cc phng php gii tng dng ton cha phn nguyn. Cc v d

minh ha phng php c la chn mang tnh cht in hnh, s lng ln

bi tp th hin s phong ph mun hnh v ca ng dng hm phn nguyn

trong gii ton v c gii chi tit trong [2] nn khng trnh by li trong

lun vn ny.

Lun vn c hon thnh di s hng dn khoa hc ca PGS TS T

Duy Phng. Xin c t lng cm n chn thnh nht ti Thy.

Tc gi xin chn cm n Trng i hc Khoa hc Thi Nguyn, ni tc

gi hon thnh chng trnh cao hc ngnh ton.

V cui cng, xin cm n gia nh, bn b v ng nghip cm thng,

ng h v gip trong sut thi gian tc gihc cao hc v vit lun vn.

H Ni, ngy 15thng 9 nm 2010

Tc gi

Nguyn Th Hng Hnh



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Chng 1

CC KIN THC C BN V PHN NGUYN

1 KHI NIM V PHN NGUYN

nh ngha 1.1 Cho mt s thc x . S nguyn ln nht khng vt qu

x c gi l phn nguyn(integer part, integral part) hay sn (floor) ca x .

Ta thng k hiu phn nguyn ca x l x . Nhiu ti liu gi phn nguyn

ca

x l snv k hiu phn nguynca x l x , v snc lin quan mt

thit vi khi nim trn x ca x . Hai khi nim trnv sn thng c

s dng trong tin hc. Trong lun vn ny ta s dng c hai k hiu phn

nguyn (sn) l x v x .

nh ngha 1.2Cho mt s thc x . S nguyn b nht khng nh hn x

c gi l trn ca x v k hiu l x .

nh ngha 1.1 v nh ngha 1.2 tng ng vi:

x z1;

.

z x z

z

0 1;

.

x z

z

v

x z 1 ;

.

z x z

z

0 1;

.

z x

z

Hn na, x x nu x v 1x x vi mi x .

nh ngha 1.3 Phn d (phn thpphn, phn l, gi tr phn- fractional

part, fractional value) ca mt s thc x , k hiu l x c nh ngha bi

cng thc x x x .



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T nh ngha 1.3 ta suy ra ngay, 0 1x vi mi x v 0z khi v

ch khi z l s nguyn.

Ta bit rng, vi mi x th tn ti s nguyn z sao cho 1z x z .

nh ngha 1.4 Gi tr nh nht gia hai s x z v 1z x c gi l

khong ccht x n s nguyn gn n nht v c k hiu l x .

Ta c 0,5x x z vi mi x .

nh ngha 1.5 S nguyn gn mt s thc x nhtc k hiu l x v

x c gi l s lm trnca x .

Khi nim lm trn s c s dng rng ri trong my tnh.

xc nh, nu c hai s nguyn cng gn x nht (ngha l khi

0,5 1 0,5x z z th z v 1z cng c khong cch ti x bng 0,5

( 1 0,5x z z x ) th ta qui c chn s ln, tc l nu 0,5z x z ,

th x z , cn nu 0,5 1z x z th 1x z .

2 CC TNH CHT C BN CA PHN NGUYN

T cc nh ngha 1.1- nhngha1.5 ta i n cc tnh cht tuy n gin

nhng rt c bn v hay s dng sau y ca phn nguyn. Cc tnh cht ny

c chng minh chi tit trong [2], v vy di y chng ti ch lit k

m khng chng minh.

Tnh cht 2.1 Vi mi x ta c

a) 1x x x hay 1x x x ;

b) 1x x x hay 1x x x .

Du bng xy ra khi v ch khi x l s nguyn.



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Tnh cht 2.2 x x x ; 0 1x ; 0 1x x x .

H qu 2.1 x z z th z v 0 1x .

Tnh cht 2.3 x z x z ; x z x vi mi z .

o li, x y th y x z vi z no .

Tnh cht 2.4Nu x th x x v 0x .

Ngc li nu x x hoc 0x th x .

Nu x l s hu t nhng khng phi l s nguyn th x cng l mt s

hu t thuc khong 0;1 .

Nu x l s v t th x cng l mt s v t thuc khong 0;1 .

Tnh cht 2.5Phn d, sn v trn c tnh cht lu ng(idempotent), tc l

khi hai ln p dng php ton th kt qu khng i:

x x ; x x v x x vi mi x .

Hn na, 0x x x vi mi x .

Nhng 0x v x x x vi mi x ;

1x , 1 1x x x x vi mi x .

Tnh cht 2.6Cc qui tc i ch (hon v), kt hp ca php ton cng v

php ton nhn; qui tc kt hp gia php ton nhn v php ton cng vn

ng cho phn nguyn v phn d.

Tnh cht 2.7 Php lm trn s x thng thng nh nu trong nh

ngha 1.5 chnh l php ly phn nguyn ca 0,5x , tc l 0,5x x .

Tnh cht 2.8Nu x y th 1x y hay 1 1x y .

Tnh cht 2.9Nu x y th x y . o li, nu x y th x y .



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Tnh cht 2.10

a) C hai sx v y l hai s nguyn khi v ch khi 0x y .

b) Trong hai sx v y c mt s nguyn v mt s khng phi l s nguyn

th 0 1x y .

c) Hai s x v y khng nguyn c tng x y l mt s nguyn khi v ch

khi 1x y .

Tnh cht 2.11aVi mi ,x y ta c

1x y x y x y ; 1x y x y x y .

Nhn xt 2.1Tnh cht2.11a c thcpht biudidngsau.

Tnh cht 2.11b

khi 0 1;

1 khi 1 2.

x y x yx y

x y x y

Tnh cht ny cng c vit di dng sau y.

Tnh cht 2.11c

khi 0 1;

1 khi 1 2.

x y x yx y

x y x y

H qu 2.2 2 2x x vi mi x .

H qu 2.3 x x v 0x x nu x ;

1x x v 1x x nu x .

H qu 2.4 x x vi mi x .

Tnh cht 2.12a Vi mi x v y l cc s thc ta c

2 2 2x y x y x y x y

v 2 2x y x y .

Nhn xt 2.2Tnh cht 2.12a c th c vit di dng sau.

Tnh cht 2.12b a) Nu 1

max ,2

x y th



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2 2 0x y x y

v 2 2 2 2x y x y x y x y .

b) Nu 1

min , max , 12

x y x y x y th

2 2 1 1x y x y

v 2 2 1 2 2 1x y x y x y x y .

c) Nu 1

min , max , 12

x y x y x y th

2 2 1x y x y

v 2 2 2 2 1x y x y x y x y .

d) Nu 1

min ,2

x y th 2 2 2 1x y x y

v 2 2 1 2 2 2x y x y x y x y .

Tnh cht 2.13Vi mi x ta lun c

1

22x x

v

1

22x x x

.

H qu 2.5Vi mi s nguyn dng ta lun c1

2 2

n nn

.

Tnh cht 2.14a Vi mi ,x y ta lun c

0x y v x y x y .

Nhn xt 2.5Tnh cht 2.14a c thpht biudidngsau y.

Tnh cht 2.14b

khi ;

1 khi .

x y y xx y

x y x y

Tnh cht 2.14c

khi ;

1 khi .

x y y xx y

x y x y



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Tnh cht 2.15Vi mi s t nhin n v vi mi s thc x ta c

1n x nx n x n .

Tnh cht 2.16Vi mi s thc x khng phi l s nguyn v vi mi snguyn n ta lun c 1x n x n .

Tnh cht 2.17Vi mi s nguyn dng n v vi mi s thc x ta lun c:

1 1

... n

x x x nxn n

.

Tnh cht 2.18Vi mi x v n l s t nhin ta lun c xx

n n

.

Tnh cht 2.19Vi mi s t nhin 3k v mi s t nhin n ta c

2 2n n n

k k k

.

Tnh cht 2.20Cho 1 2, , ..., nk k k l b n s nguyn dng. Khi y

1 21 2

...... 1nn

k k kk k k n

n

.

Tnh cht 2.21 Vi mi s nguyn kta lun c2 2k k k

.

Tnh cht 2.22 Cho , l nhng s v t dng sao cho 1 1 1

. Tp

1

, 2 , 3 , ...n na

v

1, 2 , 3 , ...n nb

to thnh mt phn

hoch ca tp s nguyn dng, tc l 1n n

a

v

1n nb

l cc tp khng giao

nhau v hp ca chng bng chnh tp tt c cc s nguyn dng.

Tnh cht di y c s dng nhiu trong tin hc.

Tnh cht 2.23Cho a v 2b l cc s t nhin bt k. Khi y log 1ba

chnh l s cc ch s ca mt s a vittrong h m c s b .



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3 HM PHN NGUYN V TH HM PHN NGUYN

T cc nh ngha phn nguyn (sn), trn, phn d, s lm trn trong 1, tac th a ra cc nh ngha sau y.

Hm snHm :f , ( ) :f x x cho tng ng mi s x vi phn

nguyn x ca n c gi l hm phn nguyn.

Trong mt s ti liu, hm phn nguyn cn c gi l hm sn (floor

function) v ngoi k hiu ( ) :f x x cn c k hiu l ( ) :f x x .

th ca hm phn nguyn

Hnh 1

Hm phn nguyn l hm hng s tng khc(nhn gi tr khng i trn tng

na khong ; 1z z vi z ); gin on loi mt ti cc im z vi

lch khng i bng 1 ( lim ( ) lim ( ) 1x z x z

f x f x

, tc l hiu gia gii hn

ca hm s khi i s x tin ti n t bn phi v t bn tri bng 1).

Nh vy, hm phn nguyn khng lin tc (gin on loi 1), nhng l na

lin tc trn. Do n l hm hng tng khc nn o hm ca n tn ti vbng 0 ti mi im khng nguyn v o hm khng tn ti (thm ch hm

s khng lin tc) ti cc im nguyn.



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Hm trnHm :f , ( ) :f x x cho tng ng mi sx vi trn

x ca n c gi l hm trn.

th ca hm trn

Hnh 2

Hm trn l hm hng s tng khc (nhn gi tr khng i trn tng nakhong ( ; 1]z z vi z );gin on loi mtti cc im x z , z vi

lch khng i bng 1 ( lim ( ) lim ( ) 1x z x z

f x f x

).

Vy, hm trn khng lin tc, nhng l na lin tc di. Do n l hm hng

tng khc nn o hm ca n tn ti v bng 0 ti mi im khng nguyn

v o hm khng tn ti ti cc im nguyn.

Mt khc, th ca hm trn c th nhn c bng cch tnh tin thhm ( ) :f x x ln trn (theo trc tung) 1 n v trn cc khong ; 1z z ,

z . Tuy nhin, ti cc im nguyn th chng nhn cc gi tr khc.

Hm phn d Hm : 0;1f t tp s thc vo tp con 0;1 ca tp

s thc , ( ) :f x x vi mi x cho tng ng mi s thc x vi phn

d x ca n c gi l hm phn d (hay hm phn phn, hm phn l).

th ca hm phn d ( )f x x x x



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Hnh 3

Hm phn d ch nhn gi tr trong na khong 0;1 , tng tng khc (tng

trn tng na khong ; 1z z vi z ) vgin on loi mtti cc im

x z , z vi lim ( ) lim ( ) 1x z x z

f x f x

. c bit, hm phn d l hm tun

honvi chu k 1, ngha l 1x x vi mi x .

Hm khong cchHm : 0;0,5f cho tng ng mi s thc x vi

khong cch ti s nguyn gn n nht c gi l hm khong ccht x ti

s nguyn gn n nht v k hiu l ( ) :f x x .

Hm khong cch ch nhn gi tr trong on 0;0,5 , tng tng khc trn

tng on

; 0,5z z v gim tng khctrn

0,5; 1z z vi z . Hm

khong cch l hm lin tc v tuyn tnh tng khc.c bit, hm khong

cch l hm tun honvi chu k 1, ngha l 1x x vi mi x .

Hm lm trn Hm :f t tp s thc vo tp s nguyn ca

tp s thc , cho tng ng mi s thc x vi s nguyn gn n nht c

gi l hm lm trnv k hiu l ( ) :f x x .

Nhn xt 3.1 Ta lun c 0,5x x vi mi x (xem Tnh cht 2.7 2). th ca hm lm trn ( ) ( ) 0,5f x x x

th ca hm ( )f x x chnh l th ca hm f x x tnh tin sang

bn tri 0,5n v (c th thy r iu ny qua so snh hai th).



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Hnh 4

T Tnh cht 2.3 2 suy ra mt tnh cht th v ca hm phn d sau y.

Tnh cht 3.1 Hm phn d v hm khong cch (t x ti s nguyn gn n

nht) l hm tun hon vi chu k nh nht bng 1.

Ta nhc li rng hm : xc nh trn tp s thc v nhn gi tr

cng trong tp s thc c gi l tun honnu tn ti mt s dng T

sao cho x T X v ( ) ( )x T x vi mi x .

S Tc gi l chu kca hm tun hon ( )x .

Hin nhin, nu ( )x l hm tun hon chu k Tth ( )x cng l hm tun

hon chu k nTvi mi s t nhin n . Tht vy, v ( )x l hm tun hon

chu k Tnn vi mi x ta c:

( ) ( ( 1) ) ( ( 1) ) ... ( )x nT x n T T x n T x .

Chng t ( )x l hm tun hon chu k nTvi mi s t nhin n .

S0 0T nh nht (nu c) trong s tt c cc chu k c gi l chu k chnh

hay chu k c sca hm tun hon ( )x .

ngn gn, khi ni hm ( )x l tun hon vi chu k T, ngi ta thng

hiu Tl chu k chnh 0T (nu c) ca ( )x .

Th d, v x n x vi mi n nn hm phn d y x c chu k l

T n vi mi n l s t nhin v chu k chnh l0 1T (xem Hnh 3).



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Tng t, v x n x vi mi n nn hm y x c chu k l T n

vi mi n l s t nhin v chu k chnh l0

1T .

Nhn xt 3.2C nhng hm tun hon khng c chu k chnh.

Th dHm Dirichlet ( )y x c nh ngha nh sau: ( ) 1y x khi x

l s hu t; ( ) 0y x khi x l s v t l mt hm tun hon c chu k l

s hu t q bt k. Tuy nhin, v tp cc s hu t khng m khng c s

nh nht (vi mi s hu t 0q ta c th tm c s2

qnh hn q cng l

s hu t) nn hm s ( )y x khng c chu k chnh, tc l khng tn ti s

0 0T sao cho 0T q vi mi chu k q (vi mi s hu t q ). Vy ( )y x

l hm tun hon khng c chu k chnh.

nh ngha Hm ( )y f x xc nh trn tp X c gi l phn tun

honchu k 0T nu vi mi x X ta c

x T X v ( ) ( )f x T f x .

Tnh cht 3.2Nu ( )y f x l phn tun hon vi chu k 0T th ( )y f x l tun hon vi chu k 2 0T . o li khng ng.



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- 16 -

Chng 2

PHN NGUYN TRONG TON S HC V IS

1 PHN NGUYN TRONG TON S HC

1.1 Mt s tnh cht b sung v s nguyn v p dng trong ton s hc

Nhiu bi ton s hc lin quan mt thit vi phn nguyn.

Ngoi cc tnh cht chung cho phn nguyn nu trong 2 Chng 1, ta cn c

mt s tnh cht khc kh th v ring cho cc s nguyn v hay c p dng

trong bi tp sau y.Chng minh cc tnh cht ny c th xem trong [2].

Tnh cht 1.1Gi s r l phn d khi chia mt s nguyn m cho mt s

nguyn dng n , m pn r vi 0,1,..., 1r n . Khi ym

r m nn

.

Tnh cht 1.2Nu p v q l nhng s nguyn dng sao chop

qkhng phi

l s nguyn th 1p pq q q

.

Tnh cht 1.3 Cho q l s t nhin, x l s thc dng bt k. C ngx

q

s t nhin khng vt qu x v chia ht cho q .

H qu 1.1Cho q v n l cc s t nhin bt k. Trong dy cc s 1, 2, ...,n

c ng nq

s chia ht cho q ; 2nq

s chia ht cho 2q ; 3nq

s chia ht

cho 3q ; ...;k

n

q

s chia ht cho kq .



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Ta nhc li, mt s t nhin bao gi cng c mt phn tch duy nht ra tha

s nguyn t, tc l 1 21 2 ... kkn p p p vi ip l cc s nguyn t khc nhau v

i l cc s t nhin.

Tnh cht 1.4(Cng thc Polignac) S m cao nht kca tha s nguyn t

q trong phn tch !n ra tha s nguyn t bng 2 3 ...n n n

kq q q

.

Th d Phn tch 6! ra tha s nguyn t: 31 2 46! 2 3 5 7 ... kkp .

Ta c 1 2 3 26 6 6 6 6

... 3 1 4

2 22 2 2

;

2 2 3 2

6 6 6 6 6... 2 0 2

3 33 3 3

;

3 2 3 2

6 6 6 6 6... 1 0 1

5 55 5 5

; 4 5 ... 0 .

Vy 4 26! 2 3 5 .

Tnh cht 1.5Nu p l s nguyn t th

!

! !k

ki

kp

pC

i p i

chia ht cho p vi

mi i tha mn iu kin 1 1ki p .

Tnh cht 1.6 (Cng thc Legendre)S cc s trongdy 1, 2, 3,...,nkhng

chia ht cho mt trong cc s nguyn t 1 2, ,..., kp p p c tnh theo cng thc

1 21 2

1 2 1 3 1 1 2 3 1 2 4 2 1

1 2

( ; , ,..., ) ...

... ...

... 1 ....

k

k

k k k k k

k

k

n n nB n p p p n

p p p

n n n n n n

p p p p p p p p p p p p p p p

n

p p p



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Th d2.1Trong dy s 1, 2,..., 32 c 9 s 1, 7,11,13,17,19, 23, 29, 31khng

chia ht cho mt trong cc s 2,3,5 . Ta c

32 32 32 32 32 32 32(32;2,3,5) 322 3 5 2.3 2.5 3.5 2.3.5

32 16 10 6 5 3 2 1 9.

B

Cc tnh cht nu trn c s dng trong mt s dng ton s hc di y.

Bi ton 1 Tm ch s tn cng ca mt s t nhin

Phng phpS dng cc tnh cht ca phn nguyn

tm ch s tn cng ca mt s ta thng s dng cc tnh cht chung vphn nguyn trong 2 Chng 1 v cc tnh cht caphn nguyn nu trn.

c bit, mt s chn chc (c tn cng bng 0) phi chia ht cho 2 v cho 5.

Th d2.2(Olympic Moscow, Vng 1, 1940)

Hi 100! c tn cng bng bao nhiu ch s 0.

GiiTheo Tnh cht 1.4, s m cao nht ca 2 v ca 5 trong phn tch 100!

ra tha s nguyn t s l:

2 3 4 5 6

100 100 100 100 100 10050 25 12 6 3 1 97

2 2 2 2 2 2

.

100 100 100

5 25 125

= 20 + 4 + 0 = 24.

Nh vy, 24 97 24 24100! 5 2 (5 2) 10k q q .

Trong phn tch s q ra tha s nguyn t khng c s 5 no nn q l s

chn nhng khng phi l s chn chc. Vy 100! c tn cng l 24 ch s 0.Th d2.3(Thi hc sinh gii bang New York, 1985. Cu hi ng i)

C bao nhiu s nguyn dng n sao cho !n c tn cng bi 25 ch s 0.

Gii !n c tn cng bi 25 ch s 0 th !n phi c phn tch ra tha s

nguyn t dng 25 25 25 25! 10 (5.2) 5 .2 .n q q q , trong q khng phi l s



20/80

- 19 -

chn chc, ngha l 25 phi l s m cao nht ca 5 trong phn tch ca !n ra

tha s nguyn t.

Theo Tnh cht 1.4, s m cao nht ca 5 trong phn tch ca !n chnh l:

2 ... 255 5 5n kn n n

S

. (*)

D thy rng vi 105n th 105 2105 105

255 5

S

. Hn na, v

104 2

104 10424

5 5S

nn 105n l s nh nht tha mn iu kin ny.

Bn s tip theo l 106, 107, 108 v 109 cng tha mn iu kin (*). Vi

110n ta c 110 2110 110

265 5

S

. Vy ch c nm s 105!, 106!, 107!,

108! v 109! c tn cng bng ng 25 ch s 0.

Bi ton 2 Ton chia ht

Phng php S dng cc tnh cht ca phn nguyn

Th d 2.4Chng minh rng ! 1.2.3...n n khng chia ht cho 2n

.GiiTheo Tnh cht 1.4, s m cao nht ca 2 trong phn tch !n ra tha s

nguyn t l: 2 ...2 2 2mn n n

k

vi 12 2m mn .

V x x vi mi x nn2 2

n n

;2 22 2

n n

;2 2m mn n

.

Cng tng v ca m bt ng thc trn ta c:

2 21 1 1 1... ... 1

2 22 2 2 2 2m m mn n nk n n n

.

Vy k n v ! 1.2.3...n n khng chia ht cho 2n .



21/80

- 20 -

Bi tp 2.1(Olympic 30.4 ln th 14, 2008. thi ngh, THPT chuyn L

Qu n) C bao nhiu s nguyn dng n khng vt qu 2008tho mn

2

n

nC khng l bi ca 4. ( k

nC l k hiu t hp chp kca n phn t)Bi tp 2.2Tm lu tha cao nht kca 7 m 1000! c th chia ht cho 7k.

Bi tp 2.3Chng minh rng 1300! chia ht cho 53169 .

Bi tp 2.4 (Thi hc sinh gii cc vng ca M, 1986. Cu hi c nhn;

Olympic 30.4 ln th 10, 2004, lp 10. thi ngh, THPT Sa ec, ng

Thp) Tm s nguyn dng nh nht Nsao cho !N chia ht cho 1212 .

Bi tp 2.5Chng minh rng nu ( 1)!n chia ht cho n th n khng phi l

s nguyn t.

Bi tp 2.6 Trong cc s t nhin t 1 n 250 c bao nhiu s khng chia

ht cho ng hai trong ba s2, 5, 7.

Bi tp 2.7 Trong cc s t nhin t 1 n 610 c bao nhiu s ng thi

khng chia ht cho 6,9,15.

Bi tp 2.8 (Thi hc sinh gii Quc gia, 1995) Tm s t nhin ln nht k

tha mn iu kin: 19951994! chia ht cho 1995k.

Th d 2.5(Thi hc sinh gii bang New York, 1986) Khi biu din trong h

m c s 8, !N c kt thc bi ng 21 ch s 0. Hy tm s nguyn

dng ln nht Nc tnh cht ny (tm biu din ca Ntrong c s 10).

GiiTrc tin ta gii thch i cht v h m c s 8.

Theo thut ton Euclid, bt k mt s t nhin n no cng u phn tch c

ra ly tha ca 8 di dng

1

1 1 08 8 ... 8

k k

k kn a a a a

, trong 0 1 1, ,..., ,k ka a a a nhn mt trong cc gi tr 0,1,2,3,4,5,6,7 v 0ka . iu

ny cho php biu din bt k s t nhin n no di dng 1 1 0...k kn a a a a

vi cc h s l mt trong 8 k t 0,1,2,3,4,5,6,7 . Biu din ny c gi l

biu din ca n trong h m c s 8.



22/80

- 21 -

tm biu din ca mt s trong c s 8 ta phi phn tch s di dng

ly tha ca 8. Th d, 10 816 20 ; 10 863 77 ; ...

mt s l chn chc trong h m c s 8, s phi c dng .8k

a trongh m c s 10. Th d, 1816 2.8 ;

28100 8 64 ;

381000 8 512 ,

By gi ta i gii bi ton.

V khi biu din trong h m c s 8, s !N c kt thc bi ng 21 ch

s 0 nn !N phi l bi ca 21 638 2 , nhng !N khng l bi ca 22 668 2 .

Ngha l, trong phn tch ca ra tha s nguyn t th s m cao nht NS ca

2 phi tha mn iu kin 63 66N

S . Theo Tnh cht 1.4 1 Chng 2, s

m cao nht ca 2 trong phn tch ca !N ra tha s nguyn t c tnh theo

cng thc 263 ... 662 2 2N kN N N

S

.

S Nnh nht tha mn bt ng thc trn l 64N v

64 2 3 4 5 6

64 64 64 64 64 6432 16 8 4 2 1 63

2 2 2 2 2 2S

,

cn

63 2 3 4 5

63 63 63 63 6331 15 7 4 1 58

2 2 2 2 2S

.

Ta c: 6364! 2 q ; 6365! 64! 65 2 65q ;

63 63 64166! 64! 65 66 2 65 66 2 2 65 33 2q q q ,

64 641 267! 64! 65 66 67 2 67 2q q ; vi 1 2, ,q q q l s l;

Nhng64 64 4 66

2 2 368! 64! 65 66 67 68 2 68 2 2 17 2q q q .

Vy, 68!c tha s 66 222 8 trong phn tch ra tha s nguyn t, hay 68!

c 22 ch s 0 trong h c s 8. Do s Nln nht m trong phn tch !N

ra tha s cha 63 212 8 , hay !N c 21 ch s 0 trong h c s 8 l 67N .



23/80

- 22 -

Bi tp 2.9 (V ch ton Bungaria, 1968) Chng minh rng knC l s l khi

v ch khi s ,k n thamn iu kin: Nu mt hng no ca s k

trong h m c s 2 l ch s 1, th cng hng ca s n trong h mc s 2 cng l ch s 1.

1.2 Nh thc Newtonv ng dng trong ton s hc cha phn nguyn

ng thc 2.1Vi mi ,a b l cc s nguyn; x l s nguyn dng khng

chnh phng; n l s t nhin, ta c th biu din n

a b x di dng

n

n na b x A B x v n

n na b x A B x ,

trong ,n nA B l cc s nguyn.

Chng minh 1Vi 2n ta c: 2

2 22 22a b x a ab x b x A B x .

Suy ra 2 22A a b x v 2 2B ab , 2A v 2B l nhng s nguyn.

Tng t, 2

2 22 22a b x a ab x b x A B x .

Theo gi thitqui npta c:

1n

n n n n n na b x A B x a b x aA bB x aB bA x

.

Vy 1n n nA aA bB x v 1n n nB aB bA l nhng s nguyn.

V ta cng c:

1

1 1

.

n

n n n n n n

n n

a b x A B x a b x aA bB x aB bA x

A B x

Vy ng thc 2.1 c chng minh.

ng thc 2.2

a) 2n

n nx y A B xy v 2n

n nx y A B xy .



24/80

- 23 -

b) 2 1n

n nx y A x B y

v 2 1n

n nx y A x B y

.

Chng minha) Vi 1n :

2

1 12 2x y x x y y x y xy A B xy .

Suy ra 1A x y v 1 2B , 1A v 1B l nhng s nguyn.

Tng t,

2

1 12 2x y x x y y x y xy A B xy .

Theo gi thitquy np ta c:

2( 1) 2 2

1 1

2

( ) 2 2 ( ) ,

n n

n n

n n n n n n

x y x y x y A B xy x y xy

A x y B xy A B x y xy A B xy

1 2n n nA A x y B xy v 1 2n n nB A B x y l nhng s nguyn.

Tng t,

2( 1) 2 2

1 1

2

2 2 ( ) .

n n

n n

n n n n n n

x y x y x y A B xy x y xy

A x y B xy A B x y xy A B xy

b) Vi 1n ta c:

3 3 2 2 3

1 1

3 3

3 3 3 3 .

x y x x y x y y

x x x y x y y y x y x y x y A x B y

Suy ra: 1 3A x y v 1 3B y x , 1A v 1B l nhng s nguyn.

Tng t,

3 3 2 2 3

1 1

3 3

3 3 3 3 .

x y x x y x y y

x x x y x y y y x y x y x y A x B y

Theo gi thit quy np ta c:



25/80

- 24 -

2 3 2 1 2

1 1

2

2 2 ,

n n

n n

n n n n n n

x y x y x y A x B y x y xy

A x y x A x y B x y y B y x A x B y

trong

1 2n n nA A x y B y v 1 2n n nB A x B x y

l nhng s nguyn. Tng t,

2 3 2 1 2

1 1

2

2 2 .

n n

n n

n n n n n n

x y x y x y A x B y x y xy

A x y x A x y B x y y B y x A x B y

Vy cc cng thc trong ng thc 2.2 c chng minh.

Nh thc Newton (cc ng thc 2.1 v 2.2) c p dng rt hiu qu v o

nhiu bi ton, trong c cc bi ton s hc.

Phng php 2 p dng nh thc Newton

Th d 2.6Cho 5 2 6a ; 5 2 6b .t n nnS a b .

a) Chng minh: 2 210 1; 10 1a a b b .

b) Chng minh 4nS v nS l cc s nguyn c cng ch s tn cng.

c) Tm ch s hng n v ca 48

3 2

.

d) Tm ch s tn cng ca 250

3 2

.

Giia) Ta c: 2

2 5 2 6 49 20 6 10 5 2 6 1 10 1a a .

Tng t, 2

2

5 2 6 50 20 6 1 10 5 2 6 1 10 1b b .

Nhn xt rng, a v b l hai nghim ca phng trnh 2 10 1 0x x .

b) Ta c 2 10 1a a v 2 10 1b b .

Suy ra 2 110n n na a a v 2 110n n nb b b .

Vy 2 2 1 110 10n n n n n na b a b a b hay 2 110n n nS S S .



26/80

- 25 -

Thay n bng 2n ta c 4 3 210n n nS S S . Suy ra

4 3 2 3 1 3 110 10 10 10n n n n n n n n nS S S S S S S S S

hay 4 3 110( )n n n nS S S S chia ht cho 10.

Vy 4nS v nS l hai s nguyn c cng ch s tn cng.

c) Ta c 0 0

0 5 6 5 6 2S ; 1 1

1 5 6 5 6 10S

v 2448 2 24

3 2 3 2 5 2 6

.

S

24 24

24 5 2 6 5 2 6S l s nguyn c ch s tn cng bng ch

s tn cng ca 0S v

24 0 24 20 20 16 4 0...S S S S S S S S .

Mi s hng 4 4 4n nS S vi 0,1, 2, ..., 5n u chia ht cho 10 nn 24 20S S

chia ht cho 10, m 0 2S nn 24S c ch s tn cng l 2.

Mt khc, 0 5 2 6 1 . Suy ra 24

5 2 6 1 . Vy

24 24 24

24 241 5 2 6 5 2 6 1 5 2 6S S .

Chng t 24

245 2 6 1S

. V 24S c ch s tn cng l 2 nn

48 24

243 2 5 2 6 1S

c ch s tn cng l 1.

d) Ta c 250 125

3 2 5 2 6

.

V 125 1 124 4 31 nn 125S c ch s tn cng l 0 (trng vi ch s tn

cng ca 1S). Mt khc, 0 5 2 6 1 nn 250

0 5 2 6 1 . Vy

125 125 125

125 1251 5 2 6 5 2 6 1 5 2 6S S .



27/80

- 26 -

Chng t 125

1255 2 6 1S

. M 125S c ch s tn cng l 0.

Vy s 250 125

3 2 5 2 6

c s tn cng l 9.

Th d 2.7(Tp ch Ton hc v Tui tr, thng 2, 2005) Tm s nguyn t p

nh nht 2

3 1n

p

chia ht cho 12n vi mi s t nhin n .

Gii Vi 2p , chn 2n th 4

3 2 1 378

khng chia ht cho 32 .

Vi 3p , chn 1n th 2

3 3 1 23

khng chia ht cho2

2 .

Nh vy, s nguyn t nh nht tha mn u bi ch c th 5p .

Vi 5p ta c: 2

1 3 5 14 6 5x v 2

2 3 5 14 6 5x . Vy

2 2

3 5 3 5 28 v 2 2

3 5 . 3 5 16 hay 1,2x l nghim ca

phng trnh bc hai 2 28 16 0x x . t 1 2n n

nS x x ta c:

2 2 1 12 1 2 1 2 1 2 1 2 1 2 128 16n n n n n nn n nS x x x x x x x x x x S S .

Nh vy,nS l nghim ca phng trnh sai phn cp hai

2 128 16 0n n nS S S . Do 2

20 3 5 14 6 5 1x nn 20 1n

x .

Suy ra 1 2 1n n

n n nS x S x S hay 1 1n

nx S .

Ta c 1 28S chia ht cho22 4 . Gi s nS chia ht cho

12n v 1nS chia

ht cho 22n . Khi y 2 31 2 1 1 21 28 16 2 7 2n n

n n nx S S S q q

chia

ht cho 32n hay 2

13 5 1 1n

nx chia ht cho 12n vi mi n .

Bi tp 2.10 Tm hai ch s tn cng ca s 2010

29 21

.



28/80

- 27 -

Bi tp2.11 Chng minh s 7

8 3 7 c by ch s 9 lin sau du phy.

Bi tp 2.12 (Tp ch Ton hc v Tui tr) Chng minh rng trong biu din

thp phn ca s 7 4 3 n

vi mi s t nhin 1n , c t nht n ch s 9

ngay sau du phy.

Bi tp 2.13(Tp ch Ton hc v Tui tr) Chng minh rng phn thp phn

ca 5 2 6 n

, vi mi s t nhin 1n , bt u bng n ch s ging nhau.

Bi tp 2.14Tm s m cao nht ca 2 trong phn tch 1 3 n

, n

thnh tch cc tha s nguyn t.

Bi tp 2.15 (Olympic 30.4 ln th 7, 2001, Lp 10. thi ngh, THPT

chuyn Tr Vinh) Tm s kln nht sao cho 2001

1 3

chia ht cho 2k.

Bi tp 2.16(Olympic 30.4 ln th 15, 2009, lp 10. thi ngh, Quc

hc Hu) Cho 4 15 n

nx

vi n . Tm s d ca nx khi chia cho 8 .

Bi tp 2.17(Olympic 30.4 ln th 10, 2004. thi ngh, THPT L T

Trng, Cn Th) Chng minh rng 2

3 2 n

khng chia ht cho 5 vi

mi s t nhin n .

2 TNH GI TR CA MT S HOC MT BIU THC

CHA PHN NGUYN

Tnh gi tr ca mt s hoc mt biu thc cha phn nguyn l mt dng

ton c lp, ng thi n lin quan mt thit v c th h tr gii cc

dng ton khc (chng minh h thc cha phn nguyn; gii phng trnh v

h phng trnh cha phn nguyn;...).



29/80

- 28 -

tnh mt s hocmt biu thc cha phn nguyn, ta cn s dng cc tnh

cht ca phn nguyn nu trong 2 Chng 1, kt hp vi cc k thut tnh

ton khc, c bit l:Phng php kp

nh gi s hng kp s cn tnh phn nguyn gia hai s nguyn lin

tip: abiu thc v dng 1z A z v kt lun A z ;

Di y l mt s v d minh ha.

Th d2.8(Tp ch Ton hc trong nh trng, 1981)

Tm phn nguyn ca 6 6 6 ... 6nn

a (n du cn).

GiiDy na l dy tng v 1 6 2na a vi mi n .

Ta li c: 1 6 2,449489742 3a . Vy 1 2 3a .

Ta c 2 16 6 6 6 3 3a a .

Theo quy np ta c 2 3na vi mi n . Suy ra 2na vi mi n .

Th d2.9Tm phn nguyn ca s 3 3 36 6 ... 6n

n

b

.

Gii Dy nb l dy tng v3

1 6 1,8171205928 1nb b vi mi n .

Ta c: 31 6 1,8171205928 2b ;3 3 3 3

2 6 6 6 2 8 2b .

Theo quy np ta c 3 3 3 3 3311

6 6 ... 6 6 6 2 8 2n n

n

b b

.

Vy 1 2nb vi mi n hay 1nb vi mi n .

Th d 2.10 (Tp ch Ton hc trong nh trng, 1986) Tm phn nguyn

ca s 3 3 36 6 6 ... 6 6 6 ... 6nnn

c

.



30/80

- 29 -

GiiTa c: 4 3 2 5n n nc a b vi mi n . Suy ra 4nc vi mi n .

Th d2.11Tm phn nguyn ca 2 24 16 8 3n

A n n n .

Gii Ta c: 2 22 1 4 16 8 3 2 1n n n n n .

V 2 1n v 2 2n l hai s t nhin lin tip nn 2 1nA n vi mi n .

Th d 2.12 (Thi hc sinh gii cc vng ca M, 1987) Bit phng trnh

4 3( ) 3 6 0f x x x c ng hai nghim thc 1x v 2x . Tnh 1x v 2x .

Gii Ta c 12 1x v 23 4x . Suy ra 1 2x v 2 3x .

Th d2.13Tnh phn nguyn ca ( 1)( 2)( 3)A n n n n .

GiiTa c 2 23 3 1n n A n n hay 2 3A x n n .

Th d2.14Tnh m n

, vi ,m n l cc s nguyn dng v l s v

t, n .

GiiDo l s v t nn m cng l s v t. Theo nh ngha phn nguyn

ta c , 0 1m m m m

. Vy ( )m n m m n m n m n

mn mn

.

Do 0n nn 1 0n

v ta c:

0 1m 1 0m 0n n m

0n mn

.

Suy ra 1 0n m

hay 1m n

. Vy 1m n mn

.

Th d2.15 (Tp ch Ton hc v Tui tr, s 364, 2007) Tnh S vi

432 1 3 1 4 1 1

...2 3 4

nn

Sn

.



31/80

- 30 -

GiiNhn xt rng mi s hng trong tng Su ln hn 1nn

432 1 3 1 4 1 1

... 12 3 4

nn

S nn

.

Mt khc, ta li cbt ng thc 1 1 k

kx x hay 1 1k kx x vi mi

0x . Chn 21

xk

, ta c2

1 1 11 1 1k k k

kkx

k k k

.

p dng bt ng thc trn vi 2,3,...,k n v cng li, ta c

2 2

1 1( 1) ...

2S n

n

1 1 1( 1) ...

1.2 2.3 ( 1)n

n n

1 1 1 1 1 1 11 1 ... ( 1) 1

2 2 3 1n n n n

n n n n

.

Suy ra 1n S n . Vy 1S n .

Phng php s dng nh thc Newton

Th d 2.16 (Olympic 30.4 ln th 9, 2003, lp 10. thi nghi, THPT

Trng Vng, thnh ph H Ch Minh)

Chng minh rng 2 3 n

l mt s l vi mi n .

Chng minh Theo ng thc 2.1 tn ti hai s nguyn ,A B tho mn

2 3 3n

A B v 2 3 3n

A B .

Suy ra 2 3 2 3 2n n

A .

Mt khc, v 0 2 3 1 nn 0 2 3 1n

, hay 1 2 3 0n

.

Cng v vi v hai bt ng thc trn ta c: 2 1 2 3 2n

A A .

Suy ra 2 3 2 1n

A

l mt s l vi mi n .



32/80

- 31 -

Bi tp 2.18(Tp ch Ton hc v Tui tr) Tnh 7

4 15

Bi tp 2.19(Olympic 30.4 ln th 7, 2001, lp 11. thi ngh, THPT

Thoi Ngc Hu, An Giang) Tnh 2001

45 2001

.

3 CHNG MINH CC H THC CHA PHN NGUYN

Chng minh cc h thc cha phn nguyn thc cht c th coi l chng

minh cc tnh cht ca phn nguyn. chng minh cc h thc cha phn

nguyn ta phi s dng cc tnh cht nu trong 2 Chng 1, kt hp vi

cc k thut tnh ton i s khc s dng c phng php kp. Nhng

bi tp chng minh cc h thc cha phn nguyn cng lin quan cht ch vi

cc bi tp tnh gi tr ca biu thc cha phn nguyn. Di y l cc v d.

Th d 2.17 (Cuc thi mang tn Niels Henrik Abel, 1995-1996, vng chung

kt) Cho n l s t nhin. Chng minh rng: 4 1 4 2n n .

Chng minht 4 2n a ; 4 1n b . R rng a b .

Do 4 2n a nn 0 4 2a n hay2 4 2a n .

Gi s 2 4 2a n . Khi y 2 2(mod4)a . V l v mt s a ch c th c

dng 4a k ; 4 1a k ; 4 2a k hoc 4 3a k , tc l 2a ch c th chia

ht cho 4 (khi 4a k hoc 4 2a k ) hoc chia cho 4 d1 (khi 4 1a k

hoc 4 3a k ). Suy ra 2 4 2a n hay 2 4 1a n , tc l 4 1a n . Do

a l s nguyn nn 4 1a n . Vya b . Do a b .

Bi tp2.20 (Olympic o, 1974; thi chn i tuyn Hng Kng d thi

Quc t 1988) Chng minh rng: 1 4 2n n n .



33/80

- 32 -

Bi tp 2.21 (Olympic Canada, 1987) Cho n l s t nhin. Chng minh

rng: 1 4 1 4 2 4 3n n n n n .

Bi tp2.22 (Tp ch Kvant, M532)Cho 1na n n v 4 2nb n ,

n l s t nhin. Chng minh rng:a) n na b ; b)1

016

n nb a

n n .

Bi tp 2.23a) Chng minh rng1 3 1

2 4 2n n

.

b) Tm x 1 1

2 2n x n

ng vi mi s n nguyn dng.

4 GII PHNG TRNH V H PHNG TRNH

CHA PHN NGUYN

Phng trnh v h phng trnh cha phn nguyn c mt lng bi ton rt a

dng v phong ph. Gii mi phng trnh thng i hi mt s suy lun c bit.

Tng t nh phng trnh cha tr tuyt i, gii phng trnh cha phn nguyn

thng a v gii h bt phng trnh, nhng kh hn v c thm iu kinnguyn (n hoc biu thc cha n l s nguyn). Phng trnh cha phn nguyn

cng kh hn phng trnh v t v phng trnh v t thng tng ng vi mt

h phng trnh v bt phng trnh trn tp s thc m khng i hi iu kin

nguyn. Trong ny chng ti trnh by mt sphn loi cc dng phng

trnh v cc phng php gii chung. Cc th d minh ha c chn la

nhm lm sng t phng php. Cc bi tp nu trong chng ny nhm minh

ha s lng phong ph v a dng ca phng trnh cha phn nguyn, ligii chi tit cc bi tp c trnh by trong [2].

4.1 PHNG TRNH CHA MT DU PHN NGUYN DNG

( ) ( )f x x

Phng php 1S dng nh ngha



34/80

- 33 -

( ) ( )f x x( ) ( ) 1 ( );

( ) .

x f x x

x

Nh vy, gii phng trnh ( ) ( )f x x , ta a n v gii h bt

phng trnh

( ) ( ) 0;

( ) ( ) 1;

( ) .

f x x

f x x

x

Trong nhng bi ton c th, ta thng ch gii h bt phng trnh

( ) ( ) 0;

( ) ( ) 1.

f x x

f x x

(*)

c nghim, sau kim tra iu kin ( )x vi cc nghim tm

c. Ngc li, trong mt s bi tp, nhiu khi ta li gii phng trnh

( )x z vi z l s nguyn c x hoc x (thng l tp hu hn cc

s nguyn), sau th vo h (*) c nghim.

Nhn xt 4.1Tng t nh phng trnh cha tr tuyt i, nhiu phng

trnh cha phn nguyn c nhiu nghim, thm ch c v s nghim hoc

nghim l c mt on no ca ng thng thc.

Th d 2.18Gii phng trnh 3 2 3,6x x .

Cch gii 1Ta c:

3 2 3,6x x 2 3,6

3

xx

2 3,6 2 3,61; (1)

3 32 3,6

. (2)3

x xx

x

Gii (1): (1) 3,6 3 3,6x 3,6 6,6x .

Do 3,6 6,6x nn2 3,6

3,6 5,63

x . V

2 3,6

3

x nn

2 3,6

3

xch

c th bng 4 hoc 5.



35/80

- 34 -

Nu2 3,6

43

x th 4,2x (tha mn iu kin 3,6 6,6x );

Nu2 3,6

53

x

th 5,7x (tha mn iu kin 3,6 6,6x );.

Cch gii 2Gii (2):

(2) 2 3,6

3

x hay

21,2

3

xz vi z l mt s nguyn. Hai s

2

3

xv

1,2 khng phi l s nguyn c tng l mt s nguyn nn theo tnh cht 2.10

2 Chng 1 ta c 2

1,2 13

x

hay

20,8

3

x

.

Ta c: (1) 3,6 3 3,6x 3,6 6,6x .

V 3,6 6,6x nn 2,4

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