P DNG NH THC NEWTON CHNG MINH H THC V TNH T HP

Nguyn Vn Nm L Hong Nam

I. Thun nh thc Newton

Du hiu nhn bit: Khi cc s hng ca tng c dng th ta

s dng trc tip nh thc Newton: . Vic cn li ch l kho lo chn a,b

V d 1.1: Tnh tng

D dng thy tng trn c dng nh du hiu nu trn. Ta s chn . Khi tng trn s bng

V d 1.2: Chng minh rng:

Tng t nh trn, ta ngh ngay n vic dng nh thc vi :

Nhng tng cn tm ch cha cc s hng c vi k chn nn ta phi trit tiu c cc s hng l bng cch tnh tng khc vi

Do tng cn tm l

Bi tp tng t:

1. Chng minh rng:

2. Chng minh rng:

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II.S dng o hm cp 1,2

1. o hm cp 1

Du hiu: Khi h s ng trc t hp tng dn hoc gim dn t hay tc s hng c dng hoc th

ta c th dng o hm cp 1 n tnh.

C th

Ly o hm hai v theo x ta c :

(1)

n y thay x,a bng hng s thch hp ta c tng cn tm

V d II.1: Tnh tng

Gii

Ta thy tng cn tnh c dng nh VP (1). Vic cn li ch cn chn ta tnh c tng bng 0.

Cch khc: S dng ng thc ta c tng bng :

Dng cch ny c th trnh c dng o hm do ph hp vi cc bn 11 cha hc n o hm hoc cm thy dng cha quen o hm.

V d II.2: Tnh tng

Gii

Nhn thy h s ng trc t hp gim dn nn phi hon i v tr v :

o hm theo x:

Thay x = 2, a = 3 ta c tng bng

Cch khc: Kho lo s dng 2 ng thc ta c th trnh vic phi dng o hm phc tm:

V d II.3: Tnh tng

Gii

H s trc t hp gim dn t 2008,2007,2,1 nn dng o hm l iu d hiu:

By gi nu ly o hm th ch c trong khi trong n 2008 do ta phi nhn thm x vo ng thc trn ri mi o hm:

Thay x = 1 vo ta tm c tng l

Bi tp tng t

1. Chng minh rng

2. Tnh tng

3. Tm n bit

2. o hm cp 2

Du hiu: Khi h s ng trc t hp c dng hay hay ( khng k du ) tc c dng

hay tng qut hn th ta c th dng o hm n cp 2 tnh. Xt a thc:

Khi o hm hai v theo x ta c:

o hm ln na:

n y ta gn nh gii quyt xong bi ton ch vic thay a, b, x bi cc hng s thch hp na thi.

V d II.4: Chng minh rngS =D dng thy c VT ca ng thc trn ging gn nh hon ton VP (2) ta ch vic thay l gii quyt xong bi ton

Ch : y ch l tng cn khi trnh by vo bi kim tra hay bi thi th ta phi ghi r xt a thc ri o hm 2 ln v thay x = 1 vo mi c trn s im.

Cch khc: Ta vn c th s dng c ng thc 2 ln tnh tng trn, c th:

Tng t nh trn ta d dng tnh c tng bng cch thay x = -1 v n = 16

Hoc ta cng c th s dng n gin hn mt cht.

V d II.5: Rt gn tng sau

Gii

Vi tng nh bi trn ta xt a thc

o hm ln 1:

Nu ta tip tc o hm ln na th ch thu c 1.2, 2.3 , do thu c ta phi nhn thm hai v vi x ri mi ly o hm:

Thay x = 1 ta rt gn c tng trn thnh

Tng t khi tnh tng ta cn ch l trc t hp c mt h s ln hn k trong nn ta phi nhn vi x trc khi o hm 2 ln.

Bi tp tng t:

1.Tnh tng

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III. S dng tch phn xc nh

Du hiu: tng ca phng php ny l da vo h thc

T y d dng tm c du hiu s dng phng php ny l s

hng ca tng c dng . C th, xt tch phn

ta c th tnh bng hai cch.

Tnh trc tip:

Hoc gin tip:

Hai cch trn l nh nhau nn t ta c c:

Ty bi ton ta chn cc h s a, b, c, d, thch hp

V d III.1: CMR

Gii

Nhn vo t ca phn s d dng tm c hai cn .Tip tc mt cht ta chn tip suy ra pcm

Ch : Khi trnh by bi thi phi ghi r tch phn ri tnh bng hai cch mi c trn im.

Cch khc: Ta c th trnh khng dng tch phn bng cch p dng

ng thc: . Vic tnh ton khng nhng n gin hn m cn gim thiu c sai st khi lm bi:

thy r s hu ch ca ng thc n gin , ta xt mt v d khc. Tnh tng

R rng dng tch phn i vi bi ny gn nh l khng th nhng nu p dng ng thc th li l mt chuyn khc:

Vic cn li by gi ch l tnh tng trong ngoc vung . C rt nhiu cch tnh nn chng ta s quay li tng ny trong phn Cc phng php khc .

Tr li phn tch phn, vi vic thay a, b, c, d bng cch hng s thch hp ta c th ch ra cc bi ton phc tp hn, chng hn khi

ta c:

V d III.2: Tnh

Gii

Mi s hng ca tng c dng nn ta ngh ngay n dng tch phn. Nhng mu ca h s li l so vi trong du hiu trn l

. Do ta phi thay tch phn bng tch phn khc.

y ta chn .

D dng tm c cn trn l 2, cn di l 1. Th li:

Vic cn li by gi ch l i tnh trc tip I:

Vi tng ta xt tng sau:

Mu ca h s trc t hp gi y khng cn mu mc na m nhy

cc v mi s hng c dng nn s hng ban

u ca n trc khi ly nguyn hm l hay n y

phn no ta on ra c tch phn ban u l .

Nhng nh vy th du tr u ra ?

Tinh mt cht ta sa li c: . Vic thay cn n gin hn, y ta chn cn trn l 1, cn di l 0. Th li t cht:

Phn cn li ca bi ton l tnh tch phn :

Vi vic thay i tch phn ta c th lm ra ti t cc tng khc phc tp hn. V d

Mt s bi tng t:

1.Tnh tng bng hai cch:

( HD: tng t v d III.1 )

2.Tnh tng bng hai cch:

( HD: tng t v d III.1 )

3.Chng minh rng

( HD: tng t v d III.2 )

4.Chng minh rng

( HD: tng t v d III.2 )

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IV.Cng c s phc

tng ca phng php ny l da tnh cht c bit ca i:

vi

T , ta xt a thc

t . Ta c:

Vi ln lt l phn thc v phn o ca .V d IV.1: Rt gn .

Gii

R rng trong a thc . Mt khc ta c nn cng vic by gi ch l i tnh v

phn thc ca n chnh l tng cn tm:

.

Ta cng c th s dng (1), (3) ta tm ra trn gii nhng mt cng gii li h phng trnh 4 n v nh th th tht l git rui m li dng n dao m tru.

Tng t ta tnh c tng

V d IV.2: Tnh

Gii

Trc tin ta phi dng o hm c c h s ng trc t hp. Xt a thc:

Li nhn vi x ta c

Nhn thy chnh l phn o ca :

Do

Tng t ta dng o hm 2 ln tnh tng:

:

Gii:

Ta c:

Tng cn tnh l phn thc ca:

Bi tp tng t:

Cho khai trin . Tnh tng

a.

b.

c.

d.