phân thức hữu tỷ và các bài toán liên quan

7/22/2019 Phn thc hu t v cc bi ton lin quan

1/85

I HC THI NGUYNTRNG I HC KHOA HC

- - - - - - - - - - - - - - - - - -

Cao Th Kim Anh

PHN THC HU T

V CC BI TON LIN QUAN

LUN VN THC S TON HC

Chuyn ngnh: PHNG PHP TON S CPM s: 60.46.40

Ngi hng dn khoa hcGS. TSKH. NGUYN VN MU

THI NGUYN - NM 2011

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


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Mc lc

M u 2

1 Phn thc hu t v cc bi ton lin quan 4

1.1 Phn thc hu t v cc tnh cht lin quan . . . . . . . . . . 41.1.1 nh ngha v cc tnh cht c bn . . . . . . . . . . . 41.1.2 Phn tch phn thc hu t thnh nhn t . . . . . . 71.1.3 Mt s tnh ton trn phn thc hu t . . . . . . . . 91.1.4 Mt s lp phng trnh vi hm phn thc hu t . . 131.1.5 Phng trnh hm trong lp hm phn thc hu t . . 21

1.2 Mt s thut ton tm nguyn hm ca hm phn thc hu t 241.2.1 Phng php Oxtrogradski . . . . . . . . . . . . . . . 251.2.2 p dng cng thc ni suy Lagrange . . . . . . . . . 28

1.2.3 p dng cng thc ni suy Hermite . . . . . . . . . . 301.2.4 Phng php Horowitz . . . . . . . . . . . . . . . . . 37

2 Phn thc hu t vi h s nguyn v phn thc nhn gitr hu t 422.1 Hm phn thc chnh quy hu t . . . . . . . . . . . . . . . 42

2.1.1 Hm phn thc chnh quy hu t mt bin . . . . . . 422.1.2 Hm phn thc chnh quy hu t nhiu bin . . . . . 43

2.2 Tnh cht ca hm phn thc nhn gi tr hu t . . . . . . . 46

2.3 Dy s sinh bi hm phn thc hu t . . . . . . . . . . . . . 523 Bt ng thc vi cc hm phn thc hu t 57

3.1 Phng php bt ng thc Cauchy i vi hm phn thchu t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.2 K thut cng mu s Engel ca bt ng thc Chebyshev . 653.2.1 Bt ng thc Chebyshev v Chebyshev dng Engel . 653.2.2 Phng php cng mu s Engel . . . . . . . . . . . . 69

3.3 Dng phn thc ca cc a thc i xng c bn . . . . . . . 76Kt lun

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . 84



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M u

Phn thc hu t l mt trong nhng khi nim c bn ca chng trnhTon bc hc ph thng. c bit, cc trng THPT chuyn v cc lpchuyn ton c rt nhiu dng ton lin quan n hm phn thc. Trongcc k thi hc sinh gii Ton trong nc v cc k thi Olimpic Ton ca ccnc trn th gii, c nhiu bi ton v dy s, bt ng thc, phng trnh,bt phng trnh v h bt phng trnh ... sinh bi cc hm s dng phnthc v v th cn bit cch gii vn dng tnh c th ca biu thc phnthc cho. Hin nay cc ti liu c tnh h thng v vn ny cn chac cp nhiu.

p ng nhu cu hc tp v ging dy mn ton bc ph thng, lunvnPhn thc hu t v cc bi ton lin quannhm h thng v gii quyt

cc bi ton lin quan n phn thc hu t. Lun vn c chia ra lm bachng.

Chng 1 xt cc phn thc hu t v cc bi ton lin quan.Chng ny nu ln mt s kin thc c bn v phn thc hu t v cc

tnh cht c bn ca n, tp trung ch yu vo vic phn tch phn thc hut thnh phn thc n gin v gii thiu mt s phng php c bit sdng cng thc ni suy xy dng thut ton tm nguyn hm ca hmhu t nh phng php Oxtrogradski, p dng cng thc ni suy Lagrange,

p dng cng thc ni suy Hermite, v phng php Horowitz. Ngoi vicgii thiu cc thut ton, trong tng mc u c xy dng cc v d minhha v phn tch chi tit cc lc gii.

Trong chng ny cng xt mt s tnh ton trn cc phn thc hu tv kho st mt s lp phng trnh vi hm phn thc hu t.

Chng 2 kho st cc phn thc hu t vi h s nguyn v phn thcnhn gi tr hu t.

Xt lp cc hm phn thc hu t c bit, l lp hm phn thcchnh quy hu t. Chng minh nh l c bn v gi tr nh nht ca hm



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phn thc chnh quy hu t mt bin, hai bin v nhiu bin trn tp cc sdng.

Tip theo, xt tnh cht ca hm phn thc nhn gi tr hu t. Tng

t nh i vi s hu t, ta cng chng minh c rng mi phn thc hut nhn gi tr hu t trn tp cc s t nhin u c dng phn thc hut vi h s nguyn.

Mc cui ca Chng 2 nhm kho st mt s lp bi ton v dy s sinhbi mt s hm phn thc hu t.

Chng 3 nhm kho st mt s dng bt ng thc vi hm phn thchu t.

Trc ht xt cc k thut c bn v s dng bt ng thc Cauchy

chng minh cc bt ng thc dng phn thc.Tip theo, da vo sp th t ca dy s vn dng k thut cng mu

s Engel ca bt ng thc Chebyshev chng minh mt s dng bt ngthc c dng phn thc c bit.

Phn cui ca chng l xt mt s dng phn thc ca cc a thc ixng c bn. y l nhng dng bt ng thc loi kh cn s phi kt hpcch chng minh quy np vi cc biu din tng ng.

hon thnh lun vn ny, trc nht tc gi xin c gi li cm nsu sc ti GS.TSKH Nguyn Vn Mu dnh thi gian hng dn, chbo tn tnh gip trong sut qu trnh xy dng ti cng nh honthin lun vn.

Tip theo, tc gi cng xin gi li cm n chn thnh cc thy c c, kim tra, nh gi v cho nhng kin qu bu lun vn c y hn, phong ph hn. Qua y, tc gi cng xin c gi li cm n tiBan gim hiu, phng sau i hc, phng o to, khoa Ton - Tin Trng

HKH, i hc Thi Nguyn v cc bn ng nghip to iu kin thunli trong sut qu trnh hc tp ti trng.Tuy bn thn c nhiu c gng, song thi gian, trnh v iu kin

nghin cu cn hn ch nn lun vn kh trnh khi nhng thiu st. Rtmong c s ng gp kin ca cc thy c v cc bn ng nghip lun vn c hon thin hn.

Tc gi xin chn thnh cm n!

Thi nguyn, Thng 07 nm 2011Cao Th Kim Anh



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Chng 1

Phn thc hu t v cc bi tonlin quan

1.1 Phn thc hu t v cc tnh cht lin quan

1.1.1 nh ngha v cc tnh cht c bn

nh ngha 1.1(xem [4]).a thc bcn vi cc h s thc l hm s chobi cng thc

P(x) =anxn +an1xn1 + +a2x2 +a1x+a0

viai R, i= 0, 1, . . . , nvan= 0.nh ngha 1.2 (xem [4]). Hm s f : R R c gi l phn thc hut nu tn ti cc a thcP(x)vQ(x)sao cho

f(x) =P(x)

Q(x). (1.1)

KhiP(x)vQ(x)l cc a thc nguyn t cng nhau (khng c c chung)

th (1.1)c gi l phn thc hu t chnh tc.V sau, nhm mc tiu gii quyt cc bi ton bc THPT, ta ch xt a

thc v phn thc trn trng s thc (bin thc v vi h s thc). Nhngtrng hp c bit cn s dng s phc s c ch dn ring.

Nhng phn thc hu t dng b

(x a)n hay q(x)

[p(x)]n vin1c gi

l nhng phn thc n gin.By gi ta xt biu din mi phn thc hu t thng qua cc phn thc

hu t n gin (cc biu din km theo thut ton c th dng cho vic



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tnh cc nguyn hm ca hm phn thc hu t s c trnh by trong ccmc sau ca chng).

nh l 1.1(xem [4]).Nu hai a thcg(x), h(x)nguyn t cng nhau videg g(x) =mvdeg h(x) =nth a thc bt kf(x)videg f(x)< m + nu c th biu din c di dngf(x) =r(x)g(x) + s(x)h(x),deg r(x) 0nn g(x) = 0c ba nghim thc phn bit . Do vyf(x) = 0c

ba nghim thc phn bit.

V d 1.14.Gi sf(x)l a thc bcn 2c nnghim thc phn bitx1, x1xn.. Khi a thcg(x)bc n 1,ta lun c

(i) g(x)

f(x)=

nk=1

g(xk)

f(xk)(x xk)vi mix =xk, k= 1, 2, . . . , n ,T hy tnh tng

Ts=

nk=1

xskf(xk), 0 s n 2



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v tng

Tn1=n

k=1xn1k

f(xk)

(ii) Tnh tng

S=n1k=0

k

x k

vik = cos2k

n +i sin

2k

n .

Gii. Theo nh l Lagrange, ta c

g(x)f(x)=

nk=1

g(xk)f(xk)(x xk)

vi mi x =xk, k= 1, 2, . . . , n ,v a thcg(x)c bc n 1.Chng(x) =xs,txs =

nk=1

xskf(x)

f(xk)(x xk),suy ra

Ts=n

k=1xsk

f(xk)= 0, 0 s n 2.

Vi s= n 1,c tng

Tn1=n

k=1

xn1kf(xk)

= 1.

(ii) Txn 1 =n1k=0

(x k),suy ra

nxn

1

xn 1=n

1

k=0

1x k .

Do

S=n1k=0

k

x k = nxn

xn 1 n= n

xn 1

V d 1.15.Phn tch

f(x) = 1x(x2 + 1)(x2 + 22) . . . (x2 +n2)



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(trn R) thnh tng cc phn thc n gin. T hy tnh tng di yn

k=1

(1)k(n+k)!(n

k)!(1 +k2)

Gii. Theo nh l 1.2, ta cf(x) =

a

x+

nk=1

akx+bkx2 +k2

.

T y suy ra

f(x) = 1

(n!)2x+ 2

nk=1

(1)kx(n+k)!(n k)!(x2 +k2)

Cho x= 1,ta nhn cn

k=1

(1)k(n+k)!(n k)!(1 +k2)=

1

4(1 + 22) . . . (1 +n2)) 1

2(n!)2

V d 1.16.Cho ba s thc phn bita1, a2, a3= {0, 1, 2}.Khi c

det D= 0, viD=

1

a1

1

a2

1

a31

a1+ 1

1

a2+ 1

1

a3+ 11

a1+ 2

1

a2+ 2

1

a3+ 2

Gii. Nu nh thcdet D = 0th cc dng ca ma trnDl ph thuctuyn tnh. Khi cu0, u1, u2khng ng thi bng 0 sao cho

uo

ai+

u1

ai+ 1+

u2

ai+ 2= 0, i= 1, 2, 3.

Xt hmf(x) =

u0

ai+

u1

ai+ 1+

u2

ai+ 2=

p(x)

x(x+ 1)(x+ 2)

vi a thcp(x)c bc 2. Vf(ai) = 0nn p(ai) = 0vi i= 1, 2, 3.Vyp(x) 0hay f(x) 0khi x = {0, 1, 2}.Nu tn ti ui= 0th khix ita c ui

x+i .

Khi f(x) .Do , khix gn ic |f(x)| qu ln, mu thunvif(x)

0.iu ny ch xy ra u0 =u1 =u2 = 0,hay iu gi s l sai.

Vydet D = 0.



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V d 1.17(xem [4]).Chng minh rng

b+c+d

(b

a)(c

a)(d

a)(x

a)+

c+d+a

(c

b)(d

b)(a

b)(x

b)+

d+a+b

(d c)(a c)(b c)(x c)+ a+b+c

(a d)(b d)(c d)(x d)=

x a b c d(x a)(x b)(x c)(x d).

Gii. Ta cn chng minh

(a+b+c+d) a(a b)(a c)(a d)(a x)

+ (a+b+c+d) b

(b a)(b c)(b d)(b x)+

a+b+c+d) c(c a)(c b)(c d)(c x)+

(a+b+c+d) d(d a)(d b)(d c)(d x)

(a+b+c+d) x(x a)(x b)(x c)(x d)= 0.

p dng cng thc ni suy Lagrange vif(y) = a+b+c+d y, y1 = a,y2=b, y3=c, y4=d, y5=x,ta thu c iu phi chng minh.

1.1.5 Phng trnh hm trong lp hm phn thc hu t

Trong mc ny ta xt mt s bi ton v phng trnh hm lin quan nhm phn thc hu t.

Bi ton 1.1.Chng minh rng khng tn ti hai a thc thcf(x), g(x)tha mn iu kin

x2

+ 1 =

f(x)

g(x) .

Gii. Gi s tn ti f(x), g(x) sao cho

x2 + 1 = f(x)

g(x). Ta c th coi

(f(x), g(x)) = 1.Khi , g(x)2(x2 + 1) = f(x)2.V x2 + 1 l bt kh quynn f(x) ... x2 + 1hay f(x) = (x2 + 1)h(x).

Vyg(x)2 = (x2 + 1)2h(x)2 v nh thg(x) ... x2 + 1,mu thun vi vicchn(f(x), g(x)) = 1.



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Bi ton 1.2(xem [4]).Tm cc hm sf(x)tha mn iu kin:

fx 3x+ 1 +f

3 +x

1

x =x, x = 1. (1.2)Gii. t

x 3x+ 1

=t th x=3 +t

1 t . Khi 1.2) tr thnh

f(t) +f

t 3t+ 1

=

3 +t

1 t, t = 1. (1.3)

t 3 +x

1 x =t th x= t 3t+ 1

. Khi

(1.2) f

3 +t1 t

+f(t) = t 3

t+ 1, t = 1 (1.4)

Cng v vi v ca (1.3) v (1.4), ta thu c

3 +t

1 t+t 3t+ 1

= 2f(t) +f

t 3t+ 1

+f

3 +t

1 t

, t = 1.

Suy ra 8t

1

t2 = 2f(t) + t, t=1, hay f(t) = t

2+

4t

1

t2. Th li, ta

thy hm sf(x) = x2

+ 4x1 x2 tha mn iu kin bi ra.

Bi ton 1.3(xem [4]).Cho hm s f(x)lin tc trn Rv tha mn cciu kin

f(2010) = 2009 (1.5)

f(x).f4(x) = 1, x R, (1.6)trong fn(x) =f(f . . . (f(x))), nln f. Hy tnhf(2008).

Gii. t If = f(R). Khi , ta c xf3(x) = 1, x If. T (1.5))cho ta 2009 If. Trong (1.6) th x = 2010 th f4(2010) = 1

2009 hay

1

2009 If. Mt khc,flin tc trnRnn D:=

1

2009;2009

If. Do

f3(x) = 1

x, x D. Suy rafl n nh trnD.

Hn na,flin tc trnDnn fn iu gim trnD.

Gi sx0 Dsao chof(x0)> 1x0 .



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S dng tnh n iu gim caf(x), ta c

f(x0)> 1

x0hay f2(x0)< f

1

x0 . (1.7)Suy ra

1

x0=f3(x0)> f2

1

x0

.Do

f

1

x0

< f3

1

x0

=x0. (1.8)

T (1.7) v (1.8), ta thu cx0 > f2(x0), hay f(x0) < f3(x0) = 1

x0iu

ny mu thun vi gi thitf(x0)> 1

x0 .Chng minh tng t, ta thu c kt qu khng tn tix0 Dsao cho

f(x0)< 1

x0. Vy f(x) =

1

x, x D.

Hin nhin,2008 Dnnf(2008) = 12008

.

Bi ton 1.4(xem [4]).t S= (1; +), tm tt c cc hm sf :SStha mn cc iu kin

f[x+f(y) +xf(y)] =y +f(x) +yf(x), x, y S (1.9)f(x)

x tng trn(1;0)v(0;+). (1.10)

Gii. Nu tn tia Ssao chof(a) = 0th ta ca= 0.Tht vy, trong iu kin (1.9) chox= y =a, ta c

f[a+f(a) +af(a)] =a+f(a) +af(a),

suy ra f(a2 + 2a) =a2 + 2a (vf(a) =a).t b= a2 + 2asuy raf(b) =b. Ta xt ba trng hp

- Nu b= ath a2 + 2a= a

a = 0a = 1 (loi) a= 0

- Nu b > ath a2 + 2a > a

a > 0a < 1 (loi) a >0

V b > a >0th f(b)

b

>f(a)

a

(v f(x)

x

tng trn(0;+

)). Suy ra1 >1

(iu ny v l).



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- Nu b < ath a2 + 2a < a 1 < a

f(a)

a (v

f(x)

x tng trn(0; +)). Do vy1< 1(v l).

Vy nua Sf(a) =a th a= 0.

Trong (1.9) cho y=x, ta cf(x+f(x) +xf(x)) =x+f(x) +xf(x). Suy rax+f(x) +xf(x) = 0

hayf(x) = x1 +x

. Th li ta thyf(x) = x1 +x

tha mn cc iu kin bira.

Bi ton 1.5(xem [4]).Tm tt c cc hm sf : R+ R+ tha mn iukin

xf(xf(y)) =f(f(y)), x, y R+ (1.11)Gii . t f(1) = a, thay x bi

1

f(y) ta c

1

f(y)f(1) = f(f(y)) hay

f(f(y)) = a

f(y).

Thay xbi 1

f(x)ta c

1

f(x)f

f(y)

f(x)

=f(f(y)) =

a

f(y). Suy ra

ff(y)

f(x)

=af(x)

f(y) . (1.12)

Vit li iu kin bi ra di dngx= f(f(y))

f(xf(y)).

Theo cng thc(1.12)ta cf(x) =f

f(f(y))

f(xf(y))

=a

f(xf(y))

f(f(y)) =

a

x.

D dng kim tra hm s ny tha mn cc iu kin bi ra.Vy hm s cn tm lf(x) =

a

x, x >0vi al hng s dng.

1.2 Mt s thut ton tm nguyn hm ca hm phnthc hu t

nh l Laplace khng nh rng mi hm hu t u c nguyn hm scp. Tuy nhin, vic i tm cc nguyn hm ny khng phi l mt cngvic n gin. Ni dung ca chng ny s tm hiu mt s thut ton tmnguyn hm ca mt hm s hu t cho trc, theo th t t n gin n

phc tp.



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Ch 1.1.Gi sp(x)

q(x)l mt hm s hu t tha mn iu kin(p(x), q(x)) =

1v h s ca s hng cha bc cao nht caq(x)bng 1. p dng php

chia Euclide ta c hai a thcs(x), r(x)sao cho p(x) =q(x)s(x) +r(x)vi r(x) = 0hoc deg r < deg q. Khi , ta c

p(x)

q(x)dx =

s(x)dx+

r(x)

q(x)dx.

Vic tnh

s(x)dxl d dng, ta cn gii quyt bi ton tm

r(x)

q(x)dx

videg r 1cho ta nhng hm dng A

(x a)k1 ,cn

nhng hng thc dng Bx +C

(x2 +px+q)mdxvim >1cho ta tng ca mt



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hm hu t dng Dx+E

(x2 +px+q)m1v mt tch phn dng

Bx+C(x2 +px+q)mdx

.

Tnh li dn ta c Bx +c

(x2 +px+q)mdx= F(x)

(x2 +px+q)m1+ B

x+C(x2 +px+q)dx,

trong F(x)l a thc bc b hnm 1.Nh vy, tch phn hai v cui ny, ta c

P(x)

(x a1)k! . . . (x2 +psx+qs)hsdx=

P1(x)

(x a1)k11 . . . (x2 +psx+qs)hs1+

+ + Aix aidx+ + Bjx+Cjx2 +pjx+qj dx+. . .

Ngha l v phi ta ch cn tnh nhng tch phn vi mu s l cc lytha bc 1 ca(x ai)v ca (x2 +pj+qj)m thi. Nu bng mt cchno m ta bit c a thcP1(x)v cc hng sAi, Bj , Cjth vic tchphn thc hin c mt cch d dng. Ta c th tnh c a thcP1(x)bng cch vit n di dng h s bt nh m ta s ni r hn sau y.

Ta ch rng nu ta k hiu

Q(x) = (x a1)k1 . . . (x1 +psx+qs)h,Q1(x) = (xa1)k11 . . . (x2+psx+qs)hs1, Q2(x) = (xa1) . . . (x2+psx+qs),th d thy Q(x) = Q1(x)Q2(x)v Q1(x)chnh l c chung ln nht caQ(x)v o hm Q(x)ca n v do Q1(x)c th thu c bng thutton clt.

Ta vit

P(x)Q(x)

dx= P1(x)Q1(x)

+

P2(x)Q2(x)

dx.

P1(x), P2(x)(bc b hn bc ca mu tng ng ), c coi c cc h s btnh. o hm hai v, ta c

P(x)

Q(x)=

P1(x)Q1(x)

+

P2(x)

Q2(x).

Vic xc nh cc h s bt nh caP1(x), P2(x)c thc hin theo phngphp bit, hoc l sau khi kh mu s, ta khai trin hai v v ng nht



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h s cc s hng cng ly tha ca x, hoc l ta gn choxnhng gi trc bit.

V d 1.18.Tnh 4x4 + 4x3 + 16x2 + 12x+ 8

(x3 +x2 +x+ 1)2 dx.

D thy rng(x3 +x2 +x+ 1)2 = (x+ 1)2(x2 + 1)2.

Ta c Q1(x) =Q2(x) = (x3 +x2 +x+ 1) = (x+ 1)(x2 + 1).Ta vit

4x4 + 4x3 + 16x2 + 12x+ 8(x3 +x2 +x+ 1)2 dx

= A2x+Bx+C

(x3 +x2 +x+ 1)+

D

x+ 1dx+

Ex+F

x2 + 1 dx.

T , ly o hm hai v

4x4 + 4x3 + 16x2 + 12x+ 8

(x3 +x2 +x+ 1)2 =

Ax2 +Bx+C

x3 +x2 +x+ 1

+

D

x+ 1+

Ex+F

x2 + 1 ,

(1.13)hay

4x4 + 4x3 + 16x2 + 12x+ 8 = (2Ax+B)(x3 +x2 +x+ 1)

(Ax2 +Bx+C)(3x2 + 2x+ 1) +D(x2 + 1)(x3 +x2 +x+ 1)+(Ex+F)(x+ 1)(x3 +x2 +x+ 1). (1.14)

Ta c th khai trin hai v v cn bng h s, tuy nhin php tnh s phctp. Ta s cho hai v nhng gi tr c bit, tnh ton s n gin hn.

a thcx3 + x2 + x + 1trit tiu khi tix = 1vx = i.Ta chox 1,khi

A B+C= 6.Cho x i,sau cn bng phn thc, phn o,ta c

A C= 3, B = 1.Ba h thc trn cho ta

A= 1, B= 1, C= 4.



29/85

28

Nhn hai v (1.13) vix, x , vi ch bc ca t s ca v tri v cahng thc u v phi u b km bc ca mu s tng ng hai n v, tac

D+E= 0.

Cho x 0trong (1.14), ch B = 1, C= 4, ta cD+F = 3.

Cho x 1, ch A= 1, B = 1, C= 4,ta cD+E+F = 3.

Ba h thc cui cng cho ta

D= 0, E= 0, F = 3.

Tm liA= 1, B= 1, C= 4, D= 0, E= 0, F = 3.

T 4x4 + 4x3 + 16x2 + 12x+ 8

(x3 +x2 +x+ 1)2 dx= x

2 x+ 4x3 +x2 +x+ 1

+ 3

dx

x2 + 1

= x2 x+ 4x3 +x2 +x+ 1

+ 3arctan x+C.

Phng php Otxtrogradski mang li s tnh ton n gin hn so viphng php phn tch hm di du tch phn thnh cc phn t n gin,nht l trong trng hp khi mu sQ(x)c cha tha s(ax2 +bx+c)m

vim ln. Chng hn num= 3,th theo phng php phn tch hmdi du tch phn thnh cc phn t n gin, ta phi tnhI3qua I2, I2

quaI1(theo cng thc truy hi), cn phng php Otxtrogradski trnh ciu ny.

1.2.2 p dng cng thc ni suy Lagrange

Trong mc ny, s gii thiu mt thut ton tm nguyn hm ca hm

s hu t c dngf(x) = P(x)n

i=1

(x

xi)

trong P(x)l mt a thc,xi=xj

vi mi i =j .



30/85

29

Ta nhc li cng thc ni suy Lagrange: Chonsx1, x2, . . . , xnphn bitvnsa1, a2, . . . , anty . Khi tn ti duy nht mt a thcP(x)vi bckhng vt qun

1, tha mn iu kinP(xj) =aj, vi mij = 1, 2, . . . , n .

a thc ny c xc nh theo cng thcP(x) = nj=1

aj

i=1,i=jx xixj xi .

T cng thc trn, chng ta c mt phng php tm nguyn hm huhiu ca hm s hu t m mu thc bcnc nnghim n phn bit.

tm nguyn hm ca hm s hu t c dngf(x) = P(x)ni=1

(x xi),trong

P(x)l mt a thc vxi=xjvi mii

=j, ta tin hnh theo trnh t

nh sau.Bc 1. p dng cng thc ni suy Lagrange, ta phn tchP(x)thnh

dng tngn

j=1

aj

i=1,i=j

x xixj xi , aj =P(xj), j = 1, 2, . . . , n .

Bc 2. Thc hin php chia P(x) cho ni=1

(x

xi) ta s thu c tch

phn bt nh dng n gin.

V d 1.19(xem [3]).Tm nguyn hm ca hm s

f(x) = x2

(x 1) (x 2) (x 3) (x 4).

Gii. XtP(x) =x2 vi bn nt ni suy lx1= 1; x2= 2; x3= 3; x4= 4.

Ta c a1= 1, a2= 4, a3= 9, a4= 16.p dng cng thc ni suy Lagrange, ta c

P(x) = 16

(x 2) (x 3) (x 4 ) + 2 (x 1) (x 3) (x 4) 9

2(x 1) (x 2) (x 4) + 8

3(x 1) (x 2) (x 3) .

Rt gnf(x),ta thu c hm s sau

f(x) = 1

6 (x 1)+ 2

(x 2) 9

2 (x 3)+ 8

3 (x 4).



31/85

30

T y, ta suy ra nguyn hm ca hm s cn tm l

f(x) dx=1

6 ln |x 1| + 2 ln |x 2| 9

2ln |x 3| +8

3ln |x 4| +C,

trong Cl mt hng s.

V d 1.20.[xem [3]] Tm nguyn hm ca hm s

f(x) = 3x2 + 4x+ 1

x (x 1) (x 2) (x+ 2) .

Gii. t P(x) = 3x2 + 4x+ 1v xt bn nt ni suy l x1 = 0; x2 =1; x3= 2; x4=

2.Ta ca1= 1, a2= 8, a3= 21, a4= 5.

p dng cng thc ni suy Lagrange, ta c

P(x) = 1

2(x 1) (x 2) (x+ 1) 8

3x (x 2) (x+ 2) +

+21

8x (x 1) (x+ 2) 5

24x (x 1) (x 2) .

Rt gnf(x),ta c hm s sau.

f(x) = 1

2x

8

3 (x

1)+

21

8 (x

2)

5

24 (x+ 2).

T y, ta suy ra nguyn hm ca hm s cn tm l f(x) dx=

1

2ln |x| 8

3ln |x 1| +21

8 ln |x 2| 5

24ln |x+ 2| +C,

trong Cl mt hng s.

1.2.3 p dng cng thc ni suy Hermite

Trong mc ny, s gii thiu phng php tm nguyn hm ca hmphn thc hu t c dng

r(x)

q(x)dx,vi q(x) =

ki=1

qi(x)i v mi qi(x) (i=

1, 2, . . . , k) l a thc c h s ca hng t cao nht bng 1, qi(x)khngchnh phng,(qi(x), qj(x)) = 1vi mii =j vdeg qk(x)> 0.Ch 1.2.

1). Vi mi a thc q(x), ta u c biu din q(x) =k

i=1

qii(x), trong

mi qi(x) (i = 1, 2, . . . , k) l a thc c h s ca hng t cao nht



32/85

31

bng 1, qi(x)khng chnh phng v (qi(x), qj(x)) = 1 vi mi i= j vdeg qk(x)> 0.

2) Nup(x)v q(x)l hai a thc nguyn t cng nhau (tc(p(x), q(x)) =1) th tn ti hai a thcs(x)vt(x)sao cho

p(x)s(x) +q(x)t(x) = 1.

3) Theo cng thc ni suy Hermite (xem [3]), nu(p(x), q(x)) = 1v hs ca hng t c bc cao nht caqbng 1 vdegp(x)< deg q(x),th

i) Ta c

p(x)

q(x)

dx= c(x)

d(x)

+ a(x)

b(x)

dx, (1.15)

trong a,b,c,d l cc a thc vi deg a(x) < deg b(x) v deg c(x) 1, ta xem xt hm s di du tch phn rij(x)

qji (x)

dx.

V qi(x) khng chnh phng nn (qi, q

i) = 1. Do lun tn ti cc athc s, t K[x]sao cho sqi+ tqi = rij,trong deg(s) < deg(qi) 1vdeg(t)< deg(qi).Khi

rij

q

j

i

dx= s

q

j

1

i

dx+ tq

i

q

j

i

dx.

S dng phng php tch phn tng phn (tu= t; dv = q

i

qji

dx), ta c

t.qi(x)

qji (x)

dx= t

(j 1) qj1i (x)+

t

(j 1) qj1i (x)dx.

T y ta suy ra

rij(x)

qji (x)

dx=(t)(j1)

(j 1) qj1i (x)+

s+ t

(j1)qj1i (x)

dx.

Tip tc qu trnh trn n khij 1 = 1th vic tm nguyn hm s honthnh v khi ta thu c trng hp 1.

Tip theo, xt mt s v d minh ha.

V d 1.21.Tm nguyn hm ca hm s

f(x) =r

q =

4x9 + 21x6 + 2x3 3x2 3(x7 x+ 1)2 .

Gii. Ta c q = x7 x+ 1 v q = 7x6 1 l hai a thc nguyn tcng nhau. T suy ra tn ti hai a thcs, tsao cho qs + qs = p.Vdeg(q) = 7vdeg(q) = 6nn khng mt tnh tng qut ta c th chn athcsvtnh sau.

s= a0+a1x+a2x, t= b0+b1x+b2x2

+b3x3

.



34/85

33

T ng thcqs + qs= p, thc hin php ng nht ta thu c h phngtrnh

a2+ 7b3 = 4a1+ 7b2 = 0a0+ 7b1 = 07b0 = 21a2 b3 = 2a0+a1 b1 = 0a0 b0 = 3a1+a2 b2 = 3.

Gii h ny, ta c nghim la0= 0, a1= 0, a2= 3, b0= 3, b1= 0, b2= 0v b3= 1.T suy ra

f(x)dx=

3x2

x7 x+ 1 dx+

(x3 + 3) 7x6

1

(x7 x+ 1)2dx.p dng phng php tch phn t phn bng cch tu(x) =x3 + 3, dv =

7x6 1)(x7 x+ 1)2dx,ta tnh c

(x3 + 3) 7x6 1

(x7 x+ 1)2dx= x3 + 3

x7 x+ 1 3x2

x7 x+ 1 dx.

Do f(x)dx= x3 + 3

x7 x+ 1+C,viCl mt hng s.


f(x) =r

q=

2x+ 3

(x+ 1)3(x+ 2).

Gii. Ta c q = (x+ 1)3 (x+ 2) =3

i=1

qii = q1q22q

33,trong q1 = x+ 2;

q2= 1,vq3=x+ 1.Theo thut ton Hermite, ta vit r

qdi dng

r

q =

a

x+ 2+

b

x+ 1+

c

(x+ 1)2+

b

(x+ 1)3,

trong a, b, c, dl cc hng s. ng nht thc, ta c h phng trnh

a+b = 03a+ 4b+c = 0

3a+ 5b+ 3c+d = 2a+ 2b+ 2c+ 2d = 3.



35/85

34

Gii h ny, ta tm c nghim la= 1, b= 1, c= 1vd= 1. Do r

q =

1

x+ 2+

1x+ 1

+ 1

(x+ 1)2+

1

(x+ 1)3.

T suy ra r

qdx=

1

x+ 2dx+

1x+ 1

dx+

1

(x+ 1)2dx+

1

(x+ 1)3dx.

Vy, nguyn hm ca hm s cn tm l r

qdx= ln

x+ 2x+ 1 + 1x+ 1+ 12(x+ 1)2 = ln

x+ 2x+ 1 +2x 32(x+ 1)2 +C,

trong Cl mt hng s.

V d 1.23.Tm nguyn hm ca hm sf(x) =

r

q=

5x3 3x2 21x+ 22(x 1)3(2x 3)2 .

Gii. Ta c q= (x 1)3 (2x 3)2 =3

i=1qii =q1q

22q

33, vi q1= 1; q2= 2x 3

v q3=x 1.Theo thut ton Hermite, ta vit rq

di dng

r

q =

a

2x 3+ b

(2x 3)2 + c

x+ 1+

d

(x 1)2 + e

(x 1)3 ,trong a,b,c,d,el cc hng s. ng nht thc, ta thu c h phngtrnh

2a+ 4c = 09a+b 24c+ 4d = 515a 3b+ 33c 16d+ 4e = 311a+ 3b 30c+ 21d 12e = 213a b+ 9c 9d+ 9e = 22.

Gii h phng trnh ny ta ca= 0, b= 5, c= 0; d= 0ve= 3.Do r

q=

3

(x 1)3 + 5

(2x 3)2 .

Ta c

3

(x 1)3dx= 3

2(x 1)2 + C1v 5

(2x 3)2dx= 5

2(2x 3)+ C2,viC1, C2l cc hng s. T suy ra

f(x)dx=

32(x

1)2

+ 5

2(2x

3)

=5

2x2 + 2x + 2

(x

1)2 (2x

3)

+C,

viCl mt hng s.



36/85

35


f(x) =x5 x4 + 4x3 +x2 x+ 5

x4

2x3 + 5x2

4x+ 4

.

Gii. Ta cf(x) =x5 x4 + 4x3 +x2 x+ 5

x4 2x3 + 5x2 4x+ 4 =x+ 1 + x3 x+ 1(x2 x+ 2)2 .

t r

q =

x3 x+ 1(x2 x+ 2)2 .Ta c q =

2i=1

qii = q1q22,trong q1 = 1; q2 =

x2 x+ 2.Theo thut ton Hermite, ta vit

r

qdi dng

rq

= ax+bx2 x+ 2+ cx+d(x2 x+ 2)2 ,

trong a, b, c, dl cc hng s. ng nht hai v, ta c h phng trnh

a = 1a+b = 02a b+c = 12b+d = 1.

Gii h phng trnh ny ta c a = 1, b = 1, c =2v d =1.T suy rar

q =

x+ 1

x2 x+ 2+ 2x 1(x2 x+ 2)2 .

By gi, ta s tm x+ 1

x2 x+ 2 dx.Ta c x+ 1

x2

x+ 2

dx=1

2

2x 1x2

x+ 2

dx+3

2

1

(x 1

i

7

2 )(x+

1 +

7

2 )

dx.

T suy ra

x+ 1

x2 x+ 2 dx=1

2ln(x2 x+ 2) +3

2ln

x 1 i

7

2

x+1 +i

7

2

+C1,

viC1l hng s.



37/85

36

Tip sau y chng ta s tm 2x 1

(x2 x+ 2)2dx.Theo thut ton Hermiteta c

2x 1 =t(x2

x+ 2) + (ux+v)(2x 1),vit, u, vl cc hng s. ng nht hai v, ta c

t=8

7 ; u=

4

7; v=

97

.

T suy ra

2x 1(x

2

x+ 2)2

dx= 8

7

x2

x+ 2dx+

(4

7x 9

7)(2x 1)

(x2

x+ 2)2

dx.

Ta d dng tm c nguyn hm

87

x2 x+ 2 dx= 8

7ln

x 1 i

7

2

x+1 +i

7

2

+C2,

viC2hng s.

p dng phng php tch phn tng phn, tu= 47x 97vdv =

2x 1(x2 x+ 2)2dx,ta c

(47x 97)(2x 1)(x2 x+ 2)2 dx =

(4

7x 9

7)

x2 x+ 2 + 47

x2 x+ 2 dx

= (

4

7x 9

7)

x2 x+ 2 + 4

7ln

x 1 i

7

2

x+1 +i7

2

+C3,viC3l mt hng s.

Nh vy, ta c f(x)dx=

1

2x2 + x +

1

2ln x2 x+ 2

4x

9

7 (x2 x+ 2)+

13

14ln

2x 1 i

72x+

1 i7 + C,



38/85

37

viCl mt hng s.

Cng vi mc ch gii quyt nhng bi ton trn, tnh tch phn

(1.15), Horowitz a ra mt cch khc xc nh cc a thcd(x), b(x).T , ta c mt thut ton mi nh sau.

1.2.4 Phng php Horowitz

nh l 1.4.Chop(x)

q(x)vi(p(x), q(x)) = 1v h s ca hng t c bc cao

nht ca q(x)bng 1 v degp(x) < deg q(x). Khi , cc a thc d(x)vb(x)trong cng thc (1.15) c xc nh nh sau:

d(x) =CLN(q(x), q(x)); b(x) = q(x)

d(x)

Chng minh. Vq(x) =ki=1

qii(x)nn theo (1.16), ta c

p(x)

q(x)dx=

k

i=1i

j=1

pij(x)

qji (x)

dx.

Hn na, theo cng thc ni suy Hermite (xem [3]) th p(x)

q(x)dx=

c(x)

d(x)+

a(x)

b(x)dx, deg a(x)< deg b(x); deg c(x)


39/85

38

Gi s, ta cn tnh p(x)

q(x)dx, trong (p(x), q(x) ) = 1, degp(x) 0ng vi mix >0.

Nuf(x)vg(x)l cc hm phn thc chnh quy hu t, th vi mi cp

s nguyn dng , ,hm sh(x) :=f(x) +g(x)

cng l hm phn thc chnh quy hu t.Nuf(x)l cc hm phn thc chnh quy hu t, thh(x) := [f(x)]m, m

N.cng l hm phn thc chnh quy hu t.

2.1.2 Hm phn thc chnh quy hu t nhiu bin

nh ngha 2.3 (xem [2]). [Hm phn thc chnh quy hu t nhiu bin]Hm s f(x1, x2, . . . , xm) 0c gi l hm phn thc chnh quy hu tmbin, nu

f(x1, x2, . . . , xm) =n

k=1

akxk11 x

k22 . . . x

kmm ,



45/85

44

trong

ak N, kj Z, k= 1, 2, . . . , n; j = 1, 2, . . . , m ,a111+a221+

+ann1= 0

a112+a222+ +ann2= 0. . . . . . . . .a11n+a22n+ +annm = 0

T nh ngha trn, ta d dng kim tra tnh cht sau:

Tnh cht 2.2.Hm s f(x1, x2, . . . xm)l hm phn thc chnh quy hut khi v ch khi cc hm phn thc thnh phn caf(1, 1, . . . , xk, 1, . . . , 1)cng l cc hm phn thc chnh quy hu t.

Tip theo, ta chng minh nh l c bn sau y.

nh l 2.1.Vi mi hm phn thc chnh quyf(x1, x2, . . . xm)trn tpx1 >0, x2 >0, . . . xn >0dng

f(x1, x2, . . . , xm) =n

k=1

akxk11 x

k22 . . . x

kmm ,

trong

ak N, kj Z, k= 1, 2, . . . , n; j = 1, 2, . . . , m ,a111+a221+ +ann1= 0a112+a222+ +ann2= 0. . . . . . . . .a11n+a22n+ +annm = 0

ta u c

f(x1, x2, . . . xm) n

k=1ak.

Chng minh. chng minh nh l, ta nhc li bt ng thc gia trungbnh cng v trung bnh nhn AM-GM: Vi mi b s khng mu1, u2, . . . un,ta u c

u1+u2+ +unn

nu1u2 . . . un.Du ng thc xy ra khi v ch khi

u1=u2= =un= 1.



46/85

45

S dng kt qu ny, ta thu c

f(x1, x2, . . . , xm)

a1+a2+ +an =

n

k=1 akxk11 x

k22 . . . x

kmn

a1+a2+. . . an Du ng thc xy ra khix1=x2= =xn= 1.Nhn xt 2.1.Vi mi hm s dng

g(x) =mk=1

akxk; ak 0,

t

a1+a2+ +am = p,a11+a22+ +amm =q,

th hm sf(x) :=x

q

p g(x)

l mt hm phn thc chnh quy.

Chng minh. Tht vy, ta c

f(x) =xq

p

mk=1

akxk =

mk=1

akxk qp

v v vym

k=1ak

k q

p

=

m

k=1akk

m

k=1ak

q

p=qp q

p=qp= 0,

iu phi chng minh.T y, ta thu c mt kt qu quan trng sau y.

nh l 2.2.Mi hm dng

g(x) =mk=1

akxk ; ak 0;

u c tnh chtg(x) g(1)x qp , vi mi x >0



47/85

46

trong p= a1+a2+ +am,

q=a11+a22+

+amm.

Chng minh.Tht vy, theo nhn xt trn, ta c

f(x) =xq

p

mk=1

akxk

l mt phn thc chnh quy, nn theo nh l 1.2, th

f(x) f(1).M f(1) =g(1), nn ta c ngay iu phi chng minh.

V d 2.3.Tm gi tr nh nht ca

f(x) =x1 + 2x2 + 3x3 + + 9x9; x >0Gii.Xt

f(x)

45 =

1

45x1 +

2

45x2 +

3

45x3 +

+

9

45x9

x

145 .x

2245 . . . . . x

9945 = 1.

Suy ra f(x)

45 1 f(x) 45.Do min f(x) = 45khi v ch khix= 1.

2.2 Tnh cht ca hm phn thc nhn gi tr hut

Trong phn ny ta xt mt s tnh cht ca phn thc hu t tng t

nh i vi s hu t trong s hc.Trc tin, ta xt mt s dng ton lin quan n a thc nhn gi tr

hu t v a thc nhn gi tr nguyn.

Bi ton 2.1.Xc nh cc h s ca tam thc bc haif(x) =ax2 + bx + cf(x) Qkhi x N.Gii.Vit a thcf(x)di dng

f(x) = (x 1)(x 2)2

+(x 1) +, (2.3)



48/85

47

trong

a=

2

b= 3

2+c= = +

T (2.3), ta thu c f(x) Qkhi x Nkhi v ch khi , , Q.T suy ra iu kin cn v tam thc bc haif(x) =ax2 + bx + c Qkhi x Nl cc h sa, b, chu t.

By gi ta chuyn sang chng minh kt qu ca bi ton tng qut.

nh l 2.3. a thc

P(x) =a0xn +a1x

n1 + +antha mn iu kinP(x)Qkhi xN,iu kin cn v l cc h sca a thc cho u l s hu t.

Chng minh.Vit a thcP(x)di dng

P(x) =p0(x 1)(x 2) (x n)

n! +p1

(x 1)(x 2) (x n+ 1)(n

1)!

+ +pn,

(2.4)trong

a0=p0

n!

a1= 1 + 2 + +nn!

p0+ 1

(n 1)!p1 an =p0 p1+ pn1+pn

T (2.4), ta thu cP(x)Qkhix

Nkhi v ch khip0, p1, . . . , pn

Q.

T suy ra iu kin cn v a thcP(x) Qkhi x Nl tt ccc h s ca n l cc s hu t.

Bi ton 2.2.Xc nh cc h s ca tam thc bc haif(x) =ax2 + bx + cf(x) Zkhi x N.Gii. Vit tam thc bc haif(x)di dng

f(

x) =

(x

1)(x

2)

2 +

(x

1) +,

(2.5)



49/85

48

trong

a=

2

b= 3

2+c= = +

T (2.5), s dng tnh cht ca s hc l tch ca 2 s nguyn lin tiplun chia ht cho 2ta thu c kt lun f(x) Zvi x N khi v chkhi , , Z. T suy ra iu kin cn v tam thc bc haif(x) =ax2 +bx+c Zkhi x Nl cc h sa, b, cc dng

a=

2b= 32

+

c= = +

trong , , Z.Da vo kt qu v phng php chng minh ca bi ton ny, ta c th

gii bi ton tng qut.

Bi ton 2.3.Cho 1< nN.Xc nh cc h s ca a thc

P(x) =a0xn +a1xn1 + +an P(x) Zkhi x N.Gii. Vit a thcP(x)di dng

P(x) =p0(x 1)(x 2) (x n)

n! +p1

(x 1)(x 2) (x n+ 1)(n 1)! + +pn,

(2.6)trong

a0=p0

n!a1= 1 + 2 + +n

n! p0+ 1

(n 1)!p1 an =p0 p1+ pn1+pn

T (2.6) v s dng tnh cht ca s hc l tch cans nguyn lin tiplun chia ht chon!,ta thu c kt lun: P(x) Zvi xNkhi v ch



50/85

49

khi

a0=p0

n!

a1=

1 + 2 + +nn!

p0+ 1

(n 1)!p1

an =p0 p1+ pn1+pn

trong p0, p1, . . . , pn Z.By gi ta chuyn sang xt tnh cht ca hm phn thc hu t.

Bi ton 2.4.Chng minh rng nu hm phn thc

Q(x) := 1

cx+dQ,

x

N, (2.7)

th Q(x)c dng

Q(x) = A

Cx+D, (2.8)

trong A, C, D Z.Gii.Theo gi thit thH(x) =cx + d Q,do theo bi ton??th cch sc, dl cc s hu t. Vyc =

M1

N1vd=

M2

N2,trong M1, M2, N1, N2

l cc s nguyn. T y ta cQ(x) :=

1

cx+d=

1M1

N1x+

M2

N2

= N1N2

M1N2x+M2N1, (2.9)

pcm.


Q(x) :=ax+b

cx+dQ,

x

N, (2.10)

th Q(x)c dng

Q(x) = Ax+B

Cx+D, (2.11)

trong A, B, C, D Z.Gii. Xt Q(x) Q(1).Theo nh l phn tch phn thc hu t thnhnhn t (xem nh l 1.3), ta thu c

Q(x) Q(1)x 1 = ecx+d Q, x N, n >1.



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50

Theo kt qu ca Bi ton 2.4 th

Q(x) M3N3

x 1

= M

Cx+

D,

trong M3, N3, M , C , Dl cc s nguyn. T y, chuyn v ta thu cbiu din (2.11).


Q(x) :=ax2 +bx+c

dx+e Q, x N, (2.12)

th Q(x)c dng

Q(x) = Ax2 +Bx+CDx+E

, (2.13)

trong A, B, C, D, E Z.Gii. Xt Q(x) Q(1).Theo nh l phn tch phn thc hu t thnhnhn t (xem nh l 1.3), ta thu c

Q(x) Q(1)x 1 =

px+q

cx+d Q, x N, n >1.

Theo kt qu ca Bi ton 2.5 thQ(x) M3

N3

x 1 =M1x+N1M2x+N2

,

trong M3, N3, M2, N2, M1, N1l cc s nguyn. T y, chuyn v ta thuc biu din (2.13).


Q(x) := ax2 +bx+cdx+e

Z, x N, (2.14)

th Q(x)c dng

Q(x) =C x+DhocQ(x) =Ax2 +Bx+C Z, x N, (2.15)trong C, D Z.Gii. Nhn xt rng nud= 0th ta c ngay kt lun

Q(x) =Ax2 +Bx+C Z, x N.



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Xt trng hpd = 0.Khi ta vit (2.14) di dngQ(x) Q(0)

x =

a1x+b1dx+e

Z, x N.

Theo Bi ton 2.5 thQ(x) M

x =

A1x+B1Dx+E

,

viA1, B1,D.E,Ml cc s nguyn. Vy nn

Q(x) Mx

=A1

D +

A2

Dx+E Z, x N (2.16)

viA1, A2,D.E,Ml cc s nguyn.T y suy raA2= 0v v vy A1chia ht choD,ta thu c pcm.

Bi ton 2.8.Cho hm phn thc hu t

f(x) =P(x)

Q(x) Qx Z, (P(x)Q(x)) = 1.

Chng minh rng f(x)c th biu din c di dng phn thc ca hai

a thc vi h s nguyn.Gii. Gi s

P(x) =a0+a1x+a2x2 + +amxm,

Q(x) =b0+b1x+b2x2 + +bnxn.

Ti x = j (j = 0, 1, 2, . . . m+n)hm f(x)nhn cc gi tr hu t tngng lcj.Khi ta c h phng trnh tuyn tnh vim+n+ 2n

a0, a1, a2, . . . , am, b0, b1, b2, . . . , bn

c dng

a0 + a1k + + amkm b0ck b1ck bnckkn = 0, k= 0, 1, 2, . . . m + n.

Hai nghim ca h ny cho ta hai cp a thc tng ng P(x), Q(x) vP1(x), Q1(x)c tnh cht

P(k) ckQ(k) = 0, P1(k) ckQ1(k) = 0, k= 0, 1, 2, . . . m+n.



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52

Hai cp nghim ny cho ta a thc

g(x) =P(x)Q1(x) P1(x)Q(x), deg g(x) m+n

nhn gi tr 0 ti m+n+ 1im nn g(x)0. Do P(x), Q(x)nguyn tcng nhau nn P(x) = cP1(x), Q(x) = cQ1(x).Vy h cho ch c mtnghim vi s sai khc mt t l v nh vy tn ti ma trn cpm+n+ 1trong ma trn h s ca h phng trnh nh thc ca n khc 0 vnghim nhn c l cc s hu t, iu phi chng minh.

Nhn xt 2.2.Ta cng c th gii bi ton ny bng phng php quy npton hc da trn kt qu ca cc bi ton gii mc ny (Bi ton 2.4).

2.3 Dy s sinh bi hm phn thc hu t

Bi ton 2.9.Cho dy s thc(an)c xc nh bi :

a0= 1, an+1=1

2(an+

1

3an) vi min= 1, 2, 3, . . .

Chng minh rng vi mi s nguyn n, s An = 3

3a2n 1 l mt s chnhphng.

Gii. Trc ht, ta s chng minh rng3An+1 = 4An(An+ 3)vi mi nnguyn dng.

Tht vy, ta c

An+ 3 = 3

3a2n

1

+ 3 = 9a2n3a2n

1

nn4An(An+ 3) = 36An

9a2n3a2n 1

= 108a2n(3a2n 1)2

Mt khc, th

3An+1= 9

3a2n 1=

9

3. 122(an+ 13a2n

)2 1,

iu phi chng minh.



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53

Hn na, ta tnh c An = 9nn d dng thy rng Anlun chia htcho 3 vi mi n.

Tip theo, ta s chng minh bng quy np An

3

+ 1l s chnh phng vi

mi n.Vi n= 1, th

A1

3 + 1 = 4l mt s chnh phng.

Gi s vi n = k, mnh trn cng ng, tc l tn ti s t nhinp

sao cho Ak

3 + 1 =p2.

Ta cAk+1

3 + 1 =

4Ak(Ak+ 3)

9 + 1 = 4.

Ak

3 .(1 +

Ak

3 ) + 1 = 4p2(p2 1) + 1 =

(2p2

1)2

cng l mt s chnh phng.Suy ra An

3 + 1cng ng vin= k + 1,iu phi chng minh.

T3An+1 = 4An(An+ 3)suy ra An+1= 4An.An

3 + 1

ta cng d dng

chng minh c bng quy np rng Anl s chnh phng vi min.

Bi ton 2.10(VMO 1982).Gi savzl nhng s thc,nl s nguyndng. Nu0 a 1van+1 z 1th ta c

nk=1

z akz+ak

nk=1

1 ak1 +ak

.

Gii Nua= 1th z= 1v bt ng thc ng v hai v u bng khng.Nu a= 0th c hai v u bng 1 v bt ng thc cng ng .Xt0< a a > a2 > > an >0van+1 z 1nn c s nnguyn dng r ar+1 < z ar.Vy z

ak

z+ak = z

ak

z+ak khik > r

vz ak

z+ak = z ak

z+ak khik r

ta cz ak

z+ak 1 ak

1 +ak =

z akz+ak

1 ak

1 +ak =

2ak(z 1)(1 +ak)(z+ak)

0, k > r

z aiz+ai 1 ak

1 +ak =ai

z

z+ai 1 ak

1 +ak = 2ar+1

z

(1 +ak)(z+ai) 0, k , i r



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54

v chn i+k =r+ 1.T cc bt ng thc trn suy ra hai bt ng thcsau:

n

k=1

z ak

z+ak

n

k=1

1 ak

1 +ak

.v n

k=r+1

z ak

z+ak

n

k=r+1

1 ak

1 +ak

.

Do nhn c nk=1

z akz+ak

nk=1

1 ak1 +ak

.

Ta nhc li nh ngha v dy tun hon.

nh ngha 2.4. Dy s(un)c gi l dy s tun hon (cng tnh) nutn ti ss Nsao cho un+s=unvi min N. Ssnh vy c gi lchu k ca dy. Chu k (nguyn dng) nh nht ca mt dy s tun hon

cn c gi l chu k c s ca dy s . Trong thc hnh, chng minhmt dy s l tun hon khng nht thit phi xc nh chu k c s ca n.

T nh ngha trn ta thy ngay mt dy s l tun hon chu k1khiv ch khi dy s l dy s hng. Bi vy, ni n chu k ca mt dy stun hon, khc dy hng, hin nhin ta phi coi chu k ln hn 1.

Bi ton 2.11.Cho dy s unxc nh bi u1= 2; un+ 1 = 2 +un1

2un

vi

mi n 1.Chng minh rnga) un= 0vi minnguyn dng.b) Dy khng tun hon.

Gii. Chnl gc sao chotan = 2,khi

u1= tan ; u2= 2tan

1 tan2 = tan 2

Bng phng php quy np ta cun = tan n.Mt khcun+1=

2 +un1 2un =

2 + tan(n)

1 2tan(n)= tan + tan(n)

1 tan . tan(n).Vy

un+1= tan(n+ 1)

a) T cng thcun = tan(n)ta suy ra u2n = 2un1 U2n

.T suy ra nu

tn ti n sao choun= 0th s tn ti n l un= 0.



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55

Gi s u2s + 1 = 0 tc l 2 +u2s

1 2u2s = 0 khi u2s =2 v ta c

2 =u2s=

2us

1 u2sDo U2s +us

1 = 0,suy raukv t.

iu ny v l v theo cng thc truy hi cho trong gi thit thuklhu t. Vy un=ovi minnguyn dng.

b) Gi s dy s cho l dy s tun hon.Khi tn ti hai s nguyn dngnvk,(n > k)sao cho un=uk

haytan n = tan k (n k) = m unk = 0.Mu thun vi ktqu phn a) iu ny chng t dyun cho khng tun hon

Bi ton 2.12.Cho hm sf(x) =x tan

41

x+ tan

4

Chng minh rng hm s cho l hm tun hon vi chu k bng 3.Gii.Ta c

f(x) =x sin 4 cos 4x cos 4 + sin

4

.

Suy ra

f(f(x)) =x cos 2 + sin

2

cos

2 x sin

2

=

1

x Bin i tng t nh trn, ta cf(f(f(x))) =x.

Vy hm s cho l hm tun hon vi chu k 3.

Bi ton 2.13.Cho hm sf(x) =x tan

3 1

x+ tan

3

.Chng minh rng hm s

cho l hm tun hon vi chu k bng 6.

Gii. Ta cf(x) = x sin

3 cos

3

x cos

3+ sin

3

.Suy ra

f(f(x)) = x cos 23 + sin

23

cos23 x sin 2

3

,

f(f(f(x))) = x sin0 cos x cos + sin

=1

x ,

f(f(f(f(x)))) = x cos

3

+ sin

3x sin 3 cos 3 ,



57/85

56

f(f(f(f(f(x))))) = x sin23 cos23x cos 2

3 +x sin 2

3

,

f(f(f(f(f(f(x)))))) =

x cos + sin

cos x sin =x.

Bi ton 2.14.Gi sf(x) =

3x 1x+

3

. Xc nh g(x) =f(f . . . (f(x)) . . . ).

(nln f).

Gii.Ta vit

f(x) =

32

x 12

12

x+32

=x cos 6 sin 6x sin 6 + cos

6

Bng phng php quy np, ta d dng chng minh c

g(x) =x cos n6 sin n6x sin n6 + cos

n6



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57

Chng 3

Bt ng thc vi cc hm phnthc hu t

3.1 Phng php bt ng thc Cauchy i vi hmphn thc hu t

Trong mc ny chng ta thng s dng bt ng thc trung gian sauy.

Bi ton 3.1(xem [2]).Via, b, cl nhng s thc dng, chng minh rng

am+n +bm+n +cm+n ambn +bmcn +cman.

Gii. Ta c m am+n +nbm+n

m+n ambn (bt ng thc Cauchy vim+n

s),mbm+n +ncm+n

m+n bmcn,

mcm+n +nam+n

m+n cman.

Cng cc v tng ng ca ba bt ng thc trn, ta thu c bt ngthc cn chng minh.

c c phng php gii i vi cc bt ng thc dng phn thcn gin ny chng ta rt ra nhng nhn xt t nhng bi ton c th.

Bi ton 3.2(xem [2]).Via, b, cl cc s thc dng, chng minh rng

a5

b2+b

5

c2+ c

5

a2 a3 +b3 +c3



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Gii. Ta c :

a5

b2+ab2 2a3

b5

c2+bc2 2b3

c5

a2+ca2 2c3

a3 +b3 +c3 ab2 +bc2 +ca2

Cng bn bt ng thc trn, ta thu c bt ng thc cn chng minh.

Nhn xt 3.1. m t cc s hng ca v phi ta phi s dng cc shng c cha mu s ca cc s hng v tri nhab2, bc2, ca2.


a5

bc+

b5

ca+

c5

ab a3 +b3 +c3

Gii. Ta c:

a5

bc+abc 2a3

b5

ca+abc 2b3

c5

ab+ abc 2c3

a3 +b3 +c3 3abc

Cng bn bt ng thc trn, ta thu c bt ng thc cn chng minh.


a5

b3+

b5

c3+

c5

a3

a3

b +

b3

c +

c3

a.

Gii. Ta c a5

b3+ab 2 a

3

b.

Suy ra

a5

b3 + 2ab

a3

b +

a3

b +ab

a3

b + 2a2

.



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59

Tng t, cng c

b5

c3+ 2bc

b3

c + 2b2

c5

a3+ 2ca

c3

a + 2c2

S dng bt ng thc2(a2 +b2 +c2) 2(ab+bc+ca).Cng cc bt ng thc trn ta thu c bt ng thc cn chng minh.

Bi ton 3.5(xem [2]).Via, b, cl cc s thc dng phn bit tha mniu kin

a102b100

+b102c100

+ c102a100


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60

Gii. Ta c9a3

a+ 2b+a(a+ 2b) 6a2

9b3

b+ 2c+b(b+ 2c) 6b2

9c3

c+ 2a+ c(c+ 2a) 6c2

v kt hp vi2(a2 +b2 +c2) 2(ab+bc+ca).Cng cc bt ng thc trn, ta thu c bt ng thc cn chng minh.

Bi ton 3.7(xem [2]).Via, b, cl cc s thc dng, chng minh rnga3

(b+c)2+

b3

(c+a)2+

c3

(a+b)2

1

4(a+b+c).

Gii. Ta c8a3

(b+c)2+ (b+c) + (b+c) 6a,

8b3

(c+a)2 + (c+a) + (c+a)

6b,8c3

(a+b)2+ (a+b) + (a+b) 6c.

Cng cc bt ng thc trn, ta thu c bt ng thc cn chng minh.


a3

b(c+a)

+ b3

c(a+b)

+ c3

a(b+c)

1

2

(a+b+c).

Gii. Ta c4a3

b(c+a)+ 2b+ (c+a) 6a,

4b3

c(a+b)+ 2c+ (a+b) 6b,

4c3

a(b+c)+ 2a+ (b+c) 6c.




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a4

bc2+

b4

ca2+

c4

ab2 a+b+c.

Gii. Ta ca4

bc2+b+c+c 4a,

b4

ca2+c+a+a 4b,

c4

ab2+a+b+b 4c.



a3

(a+b)(b+c)+

b3

(b+c)(c+a)+

c3

(c+a)(a+b)

1

4(a+b+c).

Gii. Ta c:

8a3

(a+b)(b+c)+ (a+b) + (b+c) 6a

8b3

(b+c)(c+a)+ (b+c) + (c+a) 6b

8c3

(c+a)(a+b)+ (c+a) + (a+b) 6c



a2

b3+

b2

c3+

c2

a3

9

a+b+c

Gii. Ta c bt ng thc

b3

a2+

c3

b2+

a3

c2 a+b+c



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62

Suy ra:

a2

b3 +

b2

c3 +

c2

a3 =

(1

b)3

(1a

)2 +

(1

c)3

(1b

)2 +

(1

a)3

(1c

)2+

1

a+

1

b +

1

c

9

a+b+c , (pcm).

Bi ton 3.12 (xem [2]).Vi a,b,c l cc s thc dng phn bit thamn iu kina+b+c= 3abc.Chng minh rng

P = bc

a3(c+ 2b)+

ca

b3(a+ 2c)+

ab

c3(b+ 2a) 1.

Gii. Ta c bt ng thc

a3

b+ 2c+

b3

c+ 2a+

c3

a+ 2b

1

3(a2 +b2 +c2). (3.1)

Tht vy,

9a3

b+ 2c+ a(b+ 2c) 6a2

9b3

c+ 2a+ b(c+ 2a) 6b2

9c3

a+ 2b+ c(a+ 2b) 6c2

3(a2 +b2 +c2) 3(ab+bc+ca)

Cng cc bt ng thc trn, ta thu c bt ng thc (3.1). p dng (3.1),ta thu c

P =

1a3

1

b+

2

c

+

1b3

1

c+

2

a

+

1c3

1

a+

2

b

1

3

1a2

+ 1

b2+

1

c2

1

3

1ab

+ 1

bc+

1

ca

=

1

3a+b+c

abc

= 1, (pcm).



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63


a5

b2+

b5

c2+

c5

a2 ab2 +bc2 +ca2.

Gii. Ta c

a5

b2+

b5

c2+

c5

a2 a3 +b3 +c3 ab2 +bc2 +ca2, (pcm).

Tip theo, xt mt s v d lin quan n c lng mt s dng phnthc hu t c th.

Bi ton 3.14(xem [2]).Chopl mt s nguyn dng,q

(0, 1).Gi s

x [qp+1, 1]vf(x) =

pk=1

x pkx+qk

Chng minh rng

f|(x)| pk=1

1 qk1 +qk

Chng minh.Ta c0< qp+1 < qp < < q


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Ta cqj(k1) xqj(k1) +x

1 qk

1 +qk =

2(qj+1 x)(1 +qk)[x+qj(k1)]

0.

Vy nn jk=1

x qkx+qk

jk=1

1 qk1 +qk

(3.3)

T (3.2) v (3.3), ta thu cpk=1

x qkx+qk

pk=1

1 qk1 +qk

hay|f(x)|

pk=1

1 qk1 +qk

Bi ton 3.15 (xem [2]).Gi sa1, a2, . . . , an (0, 1)v nk=1

ak = 1.Vi

a (0, 1)va > akvi mik,ta lun c bt ng thc

nk=1

a aka+ak

0.Khng mt tnh tng qut, gi sx y zth x2 + 1 y2 + 1 z2 + 1.Tip tc ta c

x

x2 + 1 y

y2 + 1 (x y)(1 xy) 0.

Do 1 =x+y +z x+y 2xynn xy


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T theo 3a, ta c

1

1 + 2ab+

1

1 + 2bc+

1

1 + 2ca=

a

a+ 2abc+

b

b+ 2abc+

c

c+ 2abc 3 (a+b+c)

a+b+c+ 6abc.

Kt hp vi gi thit suy ra

3 (a+b+c)

a+b+c+ 6abc 1 a+b+c 3abc.

Vy ta c iu phi chng minh.ng thc xy ra khi v ch khi a= b = c= 1.

Bi ton 3.22(Mathlinks Contest)

.Cho cc s thc dng

a, b, c

sao cho.Chng minh

1

1 +a2 (b+c)+

1

1 +b2 (c+a)+

1

1 +c2 (a+b) 3

1 + 2abc.

Gii. Bt ng thc trn tng ng vi

1

1 + 2abc 1

1 +a2 (b+c)+

1

1 + 2abc

11 +b2 (c+a)

+ 1

1 + 2abc 1

1 +c2 (a+b) 0

a(1 bc)1 +a(3 bc)+

b(1 ac)1 +b(3 ac)+

c(1 ab)1 +c(3 ab) 0

1 bc1 +abc(3 bc)+

1 ac1 +abc(3 ac)+

1 ab1 +abc(3 ab) 0.

Khng mt tnh tng qut, gi sa b c,th

1 +abc(3 ab) 1 +abc(3 ac) 1 +abc(3 bc),

v1 bc

1 +abc(3 bc) 1 ac

1 +abc(3 ac) 1 ab

1 +abc(3 ab)(do3 =ab+bc+ca 3 3

a2b2c2 nn abc 1).

Do theo 3b, ta c

1 bc

1 +abc(3 bc)+ 1 ac

1 +abc(3 ac)+ 1 ab

1 +abc(3 ab)



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3 (1 bc+ 1 ac+ 1 ab)1 +abc(3 bc) + 1 +abc(3 ac) + 1 +abc(3 ab)= 0.

Php chng minh hon tt. ng thc xy ra khi v ch khia= b = c= 1.

Bi ton 3.23(Serbian National Olympiad 2008).Cho cc s thc dngx,y ,z sao cho x+y+z= 1. Chng minh rng

1

yz+x+1

x

+ 1

zx +y+1

y

+ 1

xy+z+1

z

2731

.

Gii. Bt ng thc cho tng ng vi

x

xyz+x2 + 1+

y

xyz+y2 + 1+

z

xyz+z2 + 1 27

31.

Khng mt tnh tng qut, gi s x y z. Khi xyz+x2 + 1xyz+y2 + 1 xyz+z2 + 1.

T gi thit ny ta cng c x

xyz+x2 + 1 y

xyz+y2 + 1.

Tht vy, bt ng thc ny tng ng vi

1

yz+x+1

x

1

zx +y+1

y

yz+x+1x zx +y+1

y (x y)

1

xy 1 +z

0.

Ch rng t iu kin1 =x+y+z x+y 2xysuy raxy 0v theo gi thit trn x y. Do bt ng thc trn l

ng.Tng t ta cng suy ra cy

xyz+y2 + 1 z

xyz+z2 + 1 x

xyz+x2 + 1 y

xyz+y2 + 1 z

xyz+z2 + 1.

Do theo 3a, ta c x

xyz+x2 + 1+

y

xyz+y2 + 1+

z

xyz+z2 + 1

xyz+x2 + 1 +xyz+y2 + 1 +xyz+z2 + 1 3 (x+y+z) = 3.



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75

Suy ra

x

xyz+x2 + 1+

y

xyz+y2 + 1+

z

xyz+z2 + 1 3

3xyz+x2 +y2 +z2 + 3.

Mt khc, t bt ng thc Schur, ta c

9xyz

x+y+z 2 (xy+yz+zx) x2 +y2 +z2 ,

suy ra3xyz 1

3(x+y+z)2 2

3

x2 +y2 +z2

.

Vy nn

3xyz+x2+y2+z2+3 13

+1

3

x2 +y2 +z2

+3 10

3+

1

9(x+y+z)2 =

31

9 .

Do x

xyz+x2 + 1+

y

xyz+y2 + 1+

z

xyz+z2 + 1 3

3xyz+x2 +y2 +z2 + 3 27

31.

Php chng minh c hon tt.

Bi ton 3.24.Cho cc s thc dnga, b, csao choab + bc + ca + abc= 4.Chng minh rng

3

1

a+

1b

+ 1

c

2 (a+ 2) (b+ 2) (c+ 2)

Gii. T iu kin bi ton ta c th ta= 2x

y+z, b=

2y

z+x, c=

2z

x+y.

V bt ng thc cn chng minh c vit li thnh

3

2

x+y

z +

y+z

x +

z+x

y

2 8(x+y+z)

2

(x+y) (y+z) (z+x).

T bt ng thc(x+y) (y+z) (z+x) 89

(x+y+z) (xy+yz+zx)

ta ch cn chng minh3

2

x+y

z +

y+z

x +

z+x

y

2 9(x+y+z)

2

xy+yz+zx,

hayx+y

z +y+z

x +z+x

y

6 (x+y+z)xy+yz+zx.



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Ta c

x+y

z +

y+z

x +

z+x

y =

x+yxz+yz

+ x+z

xy+yz+

y+zxy+zx

.

Khng mt tnh tng qut, gi sx y z. Khi xz+ yz xy + yzxy+zx, v

x+y

z

y+z

x

z+x

y .

Suy rax+yxz+yz

x+zxy+yz

y+zxy+zx

.

Do theo 3b v t bt ng thc Cauchy, th

x+

y

z +y

+z

x +z

+x

y = x

+y

xz+yz+ x

+z

xy+yz+ y

+z

xy+zx

3 (x+y+z+x+y+z)xz+yz+

xy+yz+

xy+zx

.

Php chng minh hon tt. ng thc xy ra khi v ch khia= b = c= 1.

3.3 Dng phn thc ca cc a thc i xng c bn

Bi ton 3.25(xem [4])..Cho s t nhinn(n 2)vx1, x2, . . . xn [0, 1].Chng minh bt ng thc

x1+x2+. . . xnn

11 +n(1 x1)(1 x2) . . . (1 xn) .

Gii. t1 xi=ai(i= 1, 2 . . . n).Nhn xt rng0 ai 1v bt ngthc cho tr thnh

(1 a1) + (1 a2) + + (1 an)n

11 +na1a 2 . . . an

hay(a1+a2+ +an)(1 +na1a2 . . . an) n2a1a2 . . . an.

Theo bt ng thc AM-GM th

a1+a2+ +an n na1a2 . . . an,

1 +na1a2 . . . an 1 + (n 1)a1a2 . . . an n na1a2 . . . an.



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T hai bt ng thc cui ny ta thu c iu phi chng minh.

Nhn xt rng ng thc xy ra khi v ch khi

a 1 =a2= =an= 0hay

x1=x2= =xn= 1.Bi ton 3.26 (xem [4]).Cho (a1 . . . an)l cc s thc dng. t P0 =1, Pk =Pk(a1, . . . , an); Er =Er(a1, . . . , an). Chng minh rng

Pk1Pk+1

P2k (k = 1, 2, . . . n

1).

Chng minh.Gi s

f(x, y) = (x + a1y)(x + a2y) . . . (x + any) =E0xn + E2x

n1y+ + Enyn,

Eil tng tt c cc tch s khc nhau,

Pk =

k!(n

k)!

n! Ek.

V tt c cc ai > 0 v x

y = 0 khng phi l nghim ca phng trnh

f(x, y) = 0nn x

y = 0khng phi l nghim bi trong cc phng trnh

nhn t o hm ca n.T ta c th kt lun rng cc sPidng, tc l phng trnh

Pk1

x2

+ 2Pk

xy+

Pk+1

y2

= 0nhn c t f(x, y) = 0bng cch ly vi phn lin tip theo xv y. Dophng trnh ny c nghim thc nnPk1Pk+1 P2k .

Nhn xt rng, nu ccaiu dng v khng ng thi bng nhau thta c du bt ng thc thc s.

Bi ton 3.27(xem [4]).Chng minh bt ng thc

Er1Er+1 E2r .



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Chng minh.T bt ng thc Markuc-Lopes, ta c

Pk1Pk+1 P2k .

Suy ra(k 1)!(n k+ 1)!

n! Ek1

(k+ 1)!(n k+ 1)!n!

Ek+1 0 (i

{1, . . . , n

}) phn bit.

Chng minh bt ng thc

P1 > P

1

22 > P

1

33 > > P

1

nn .

Chng minh.Ta c

P0P2 < P21 ,

(P1P3)2 < P42

(Pr1Pr+1)r < P2rr

suy ra(P0P2)(P1P3)

2 . . . (Pr1Pr+1)r < P21 P42 . . . P

2rr

Prr+1< Pr+1r P

1

r r < P

1

r+ 1r+1 .

Nhn xt 3.3.Ta c th chng minh Pr1Pr+1 < P

2r bng phng php

quy np.Gi s bt ng thc ng vin 1s dnga1, a2, . . . an1v tEr, Pr

l ccEr, Prto bin 1s y v gi s tt c cc s khng ng thibng nhau.

Khi Er =anE

r1Pr =

r

nPr+

r

nanP

r1.

T suy ran2(Pr1Pr+1 P2r) =A+Ban+C a2n,



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trong

A= {(n r)2 1}Pr1Pr+1 (n r)2PrB = (n r+ 1)(r+ 1).Pr1Pr+1+ (n r 1)(r 1)Pr2Pr+1 2(r 1)Pr2Pr+1C= (r2 1)Pr2Pr r2P2r1.V ccaikhng ng thi bng nhau nn theo gi thit, ta c

Pr1Pr+1< P

rP

r2 Pr < Pr1,

Pr2Pr+1< P

r1P

rA < PrB


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80

Bi ton 3.30(xem [4]). Choa1, . . . , an

0;1

2

ta lun c bt ng thc

di yn

k=1 ak n

k=1

ak

n n

k=1(1 ak) nk=1

(1 ak)n .

Gii. Bt ng thc cn chng minh c suy ra t v d trn vi1 = =n = 1

n.

Bi ton 3.31(xem [5]). ChoABCvi di cc cnh la, b, cv bn

knh ng trn lr.Chng minh rng(a b)2

ab +

(b c)2bc

+(c a)2

ca

p2

4r2 27

4 .

Gii. Va, b, c,l nghim ca phng trnh (xem [2],[5])

x3 2px2 + (p2 +r2 + 4Rr)x 4Rrp = 0,nn d dng kim tra h thc

(a b)2ab

+(b c)2bc

+(c a)2ca

=2(p2 +r2 + 4Rr)

4Rr 9.

Nh vy(a b)2

ab +

(b c)2bc

+(c a)2

ca =

2(p2 +r2)

4Rr 7.

Bi vR 2r,nn

(a b)2

ab

+(b c)2

bc

+(c a)2

ca

2(p2 +r2)

8r2

7 =

p2

4r2

27

4

.

Bi ton 3.32(xem [5]). ChoABCvi bn knh ng trn ni tip lrvi di ba cnh la,b,c.Chng minh rng bt ng thc

1

a2+

1

b2+

1

c2

1

4r2.

Gii. V a,b,c,l nghim ca x3 2px2 + (p2 +r2 + 4Rr)x 4Rrp = 0(xem [2],[5]) nn

f(x) =x3 2px2 + (p2 +r2 + 4Rr)x 4Rrp = (x a)(x b)(x c).



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Ly hai ln o hm ta thu c3x 2p= (x a) + (x b) + (x c).Khi

3x

2p

f(x) =

1

(x a)(x b)+ 1

(x b)(x c)+ 1

(x c)(x a) .Cho x= p,ta thu c

1

(p a)(p b)+ 1

(p b)(p c)+ 1

(p c)(p a)= 1

r2.

Vy

1

4r2 =

1

4(p a)(p b)+

1

4(p b)(p c)+

1

4(p c)(p a)

1

a2+

1

b2+

1

c2.

Bi ton 3.33.Cho a, b, cl di ba cnh ca mt tam gic khng cn.Chng minh bt ng thc

1

(a2 b2)(a2 c2)(1 +a2)+ 1

(b2 c2)(b2 a2)(1 +b2)+

1

(c2

a2)(c2

b2)(1 +c2)

1

8abc

Gii. Biu din

T(x) := 1

(x2 +a2)(x2 +b2)(x2 +c2)=

Ax+A

x2 +a2 +

Bx+B

x2 +b2 +

Cx+C

x2 +c2 .

TnhA, A, B , B, C , C v chox= 1,ta nhn c biu thc

T = 1

(1 +a2)(1 +b2)(1 +c2)

1

8abc



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Kt lun

Lun vn "Phn thc hu t v cc bi ton lin quan" t c mts kt qu sau:

- H thng ha mt s kin thc c bn v phn thc hu t v v cc

tnh cht c bn ca n, chng minh nh l phn tch phn thc hu tthnh nhn t (nh l 1.3).- Phn tch phn thc hu t thnh phn thc n gin v gii thiu mt

s phng php c bit s dng cng thc ni suy xy dng thut tontm nguyn hm ca hm hu t nh phng php Oxtrogradski, p dngcng thc ni suy Lagrange, p dng cng thc ni suy Hermite, v phngphp Horowitz.

Ngoi vic gii thiu cc thut ton, trong tng mc u c xy dng cc

v d minh ha v phn tch chi tit cc lc gii.- Xt mt s tnh ton trn cc phn thc hu t.- Kho st mt s lp phng trnh vi hm phn thc hu t.- Kho st lp cc hm phn thc hu t c bit, l lp hm phn

thc chnh quy hu t. Chng minh nh l c bn v gi tr nh nht cahm phn thc chnh quy hu t mt bin, hai bin v nhiu bin.

- Kho st cc tnh cht ca hm phn thc nhn gi tr hu t. Tng

t nh i vi s hu t, ta cng chng minh c rng mi phn thc hut nhn gi tr hu t trn tp cc s t nhin u c dng phn thc hut vi h s nguyn.

- Kho st mt s lp bi ton v dy s sinh bi mt s hm phn thchu t.

- Kho st mt s dng bt ng thc vi hm phn thc hu t. Trcht xt cc k thut c bn v s dng bt ng thc Cauchy chng minhcc bt ng thc dng phn thc.

Tip theo, da vo sp th t ca dy s vn dng k thut cng mu



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83

s Engel ca bt ng thc Chebyshev chng minh mt s dng bt ngthc c dng phn thc c bit.

Phn cui ca chng l xt mt s dng phn thc ca cc a thc i

xng c bn. y l nhng dng bt ng thc loi kh cn phi kt hpcch chng minh quy np vi cc biu din tng ng.



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Ti liu tham kho

[1] Phan c Chnh,Bt ng thc, NXB Gio Dc Vit Nam, 1993.

[2] Nguyn Vn Mu,Bt ng thc - nh l v p dng, NXB Gio Dc,2006.

[3] Nguyn Vn Mu, Cc bi ton ni suy v p dng, NXB Gio Dc,2008.

[4] Nguyn Vn Mu,a thc i s v phn thc hu t, NXB Gio Dc,2004.

[5] T Duy Phng, Phng trnh bc ba v cc bi ton trong tam giac,NXB Gio Dc, 2006.

[6] Tp ch ton hc v tui tr,Tuyn tp 5 nm, NXB Gio Dc, 2003.

[7] Ti liu t Internet.

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