Đề Tài Nckh Nguyễn Thị Thu Hà - Mông Thùy Linh - Đỗ Thị Huyền Trang

7/24/2019 Ti Nckh Nguyn Th Thu H - Mng Thy Linh - Th Huyn Trang

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I HC THI NGUYN

TRNG I HC S PHM

NGUYN TH THU H

MNG THY LINH

TH HUYN TRANG

RN LUYN K NNG GII TON BT NG THC

V BT PHNG TRNH CHO HC SINH THPT

BNG PHNG PHP HM S

TI NGHIN CU KHOA HC

Thi Nguyn 2015


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i

I HC THI NGUYN

TRNG I HC S PHM

NGUYN TH THU H

MNG THY LINH

TH HUYN TRANG

RN LUYN K NNG GII TON BT NG THC

V BT PHNG TRNH CHO HC SINH THPT

BNG PHNG PHP HM S

TI NGHIN CU KHOA HC

Ngnh: Ton

Ngi hng dn khoa hc: PGS. TS. Cao Th H

Thi Nguyn 2015


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ii

LI CM N

Trc ht, chng em xin by t lng knh trng v bit n su sc ti

PGS.TS Cao Th H- c l ngi tn tnh hng dn, ng vin v gip

chng em hon thnh ti.

Bn cnh , chng em gi li cm n chn thnh n cc thy c gio

trong t b mn l lun v phng php ging dy mn ton, Trng i hc

S phm Thi Nguyn to iu kin thun li cho chng em trong qu

trnh hc tp, nghin cu.

Cui cng, chng em xin by t lng bit n su sc ti nhng ngi

thn trong gia nh v bn b lun bn chng em, ng vin v gip

chng em hon thnh ti.

D c nhiu c gng, tuy nhin ti khng th trnh khi nhng

thiu st cn c gp , sa cha.Chng em rt mong nhn c nhng

kin, nhn xt ca cc thy c gio v bn c.

Thi Nguyn, thng 3 nm 2015

Sinh vin thc hin

Nguyn Th Thu H

Mng Thy Linh

Th Huyn Trang


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iii

MC LC

Trang

Trang ba ph .......................................................................................................... iLi cm n ............................................................................................................. ii

Mc lc ................................................................................................................. iii

Danh mc ch vit tt .............................................................................................. v

M U ................................................................................................................ 1

1. L do chn ti .................................................................................................. 1

2. Mc ch nghin cu ........................................................................................... 3

3. Nhim v nghin cu........................................................................................... 3

4. Phng php nghin cu ..................................................................................... 3

5. Khch th nghin cu v i tng nghin cu ................................................... 3

6. Gi thuyt khoa hc ............................................................................................. 3

7. D kin cu trc ca ti .................................................................................. 4

Chng 1: C S L LUN V THC TIN.................................................. 6

1.1.

K nng v k nng gii ton cho HS THPT .................................................. 6

1.1.1. K nng ......................................................................................................... 6

1.1.2. K nng gii ton ........................................................................................... 9

1.2. Hm s v quan im hm s trong dy hc ton trng THPT ................ 14

1.2.1. Khi nim hm s ........................................................................................ 14

1.2.2. Quan im hm s trong dy hc ton trng ph thng .......................... 15

1.3. Thc tin rn luyn k nng gii bi ton BT v BPT cho HS THPT

theo quan im hm s. ......................................................................................... 16

1.3.1. Thc trng vic dy hc gii BPT v BT theo quan im hm s

trng ph thng. .................................................................................................. 16

1.3.2. Nhng kh khn sai lm thng gp ca HS trong khi gii BPT v BT

theo quan im hm s .......................................................................................... 17

1.4. Kt lun chng 1 .......................................................................................... 20


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Chng 2: MT S BIN PHP RN LUYN K NNG GII BI

TON BT NG THC V BT PHNG TRNH CHO HC SINH

THNG THEO QUAN IM HM S........................................................... 21

2.1. Tng quan v bt ng thc v bt phng trnh trong chng trnh TonTHPT .................................................................................................................... 21

2.1.1. nh ngha bt ng thc v bt phng trnh .............................................. 21

2.1.2. Cc phng php thng thng gii bt phng trnh v chng minh

bt ng thc ......................................................................................................... 23

2.1.3. Cc k nng cn rn luyn cho HS khi hc ni dung ny. ............................ 27

2.2. Mt s bin php rn luyn k nng gii bi ton bt ng thc theo quan

im hm s. ......................................................................................................... 29

2.2.1. Rn luyn k nng vn dng o hm chng minh bt ng thc ............ 29

2.2.2. Rn luyn k nng s dng hm s n iu chng minh bt ng thc. ......... 44

2.2.3. Rn luyn k nng chng minh bt ng thc v tm GTLN-GTNN ca

biu thc bng phng php hm s ...................................................................... 48

2.3. Rn luyn mt s k nng gii bt phng trnh theo quan im hm. ........... 50

2.3.1. Rn luyn k nng gii bt phng trnh bng phng php vn dng

khi nim TX v tp gi tr. ................................................................................ 50

2.3.2. Rn k nng gii bt phng trnh thng qua s dng tnh n iu ca

hm s ................................................................................................................... 52

2.3.3. Rn k nng gii bt phng trnh bng phng php s dng o hm

v xt s bin thin ................................................................................................ 56

2.4. Kt lun chng 2. ......................................................................................... 59

KT LUN .......................................................................................................... 60

TI LIU THAM KHO................................................................................... 61


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v

CC CH VIT TT TRONG TI

STT Ch vit tt Ch vit y

1 BT bt ng thc.

2 BPT bt phng trnh.

3 GTLN gi tr ln nht.

4 GTNN gi tr nh nht.

5 CMR chng minh rng

6 pcm iu phi chng minh7 SGK sch gio khoa

8 THPT trung hc ph thng

9 HS hc sinh

10 TX tp xc nh


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1

M U

1. L do chn ti

Ngy nay, nc ta ang trn thc hin cng nghip ha hin i ha t

nc. Tri qua hn 20 nm i mi, nn gio dc nc ta c nhng thay i

ng k. Tuy nhin, cng cuc i mi gio dc vn lun l mt nhim v cp bch

v c t ln trn ht. Ngh quyt 29.NQ/TWv i mi cn bn v ton din

gio dc ch ra cn: tip tc i mi mnh m v ng b cc yu t c bn ca

gio dc, o to theo hng coi trng pht trin phm cht v nng lc ngi

hc. Chnh v vy, gio dc v o to lun c ng v Chnh ph ta coi l

quc sch hng u, l s nghip ca ng, Nh nc v ca ton dn. u t chogio dc l u t cho pht trin, c u tin i trc trong cc chng trnh, k

hoch pht trin kinh t-x hi.

Trong nhng nm gn y, vic nng cao cht lng dy hc ang l mt

yu cu cp bch i vi nghnh gio dc nc ta.Mt trong nhng khu then cht

thc hin yu cu ny l i mi ni dung v phng php dy hc. iu 28

khon 2 Lut Gio Dc c nu: phng php gio dc ph thng phi pht huy

tnh tch cc ch ng, sng to ca HS ph hp vi c im ca tng lp hc,

mn hc; bi dng phng php t hc, kh nng lm vic theo nhm, rn luyn

k nng vn dng kin thc vo thc tin, tc ng n tnh cm, em li nim vui,

hng th hc tp cho HS.

Mn Ton l mt mn hc c v tr c bit quan trng so vi cc mn hc

khc trng Ph thng. Bi mt mt Ton hc c ngun gc t thc tin. Mt

khc mn Ton c coi nh mn hc c s, l cng c cho nhiu mn hc khc.Trong thc t rt nhiu vn ca cc nghnh khoa hc k thut c gii quyt

nh s gip c lc ca ton hc.

Trong cc ni dung dy hc mn Ton trng ph thng, ch BPT v

BT l mt ni dung quan trng. Kin thc v k nng trong ch ny s l chic

cha kha tt nht gii quyt nhiu vn thuc hu ht cc ch kin thc v

i s, gii tch v hnh hc. Chnh v vy, bn cnh vic ging dy cc kin thc l

thuyt v ch BPT v BT mt cch y theo quy nh ca chng trnh dy


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hc th vic rn luyn k nng gii BPT v BT cho HS c ngha quan trng trong

vic nng cao cht lng dy hc mn Ton trng THPT.

Bn cnh , kin thc hm s cng chim mt v tr rt quan trng. Hm s

gi vai tr trung tm, vic bo m v tr trung tm ca hm s s tng cng tnhthng nht ca gio trnh Ton ph thng, gp phn xa b ranh gii gi to gia

cc phn khc nhau ca chng trnh.

Theo Nguyn B Kim ([9]), quan im hm c th hin trong chng

trnh Ton trng trung hc ph thng nh sau:

- Nghin cu hm s c coi l nhim v ch yu sut chng trnh bc

ph thng ;

-

Phn ln chng trnh i s v Gii tch ginh cho vic trc tip

nghin cu hm s ;

- Cp s cng v cp s nhn c nghin cu nh nhng hm s ca i

s t nhin;

- Lng gic ch yu nghin cu nhng hm s lng gic, cn phn cng

thc bin i c gim nh ;

-

Bt ng thc v bt phng trnh c trnh by lin h cht ch vi

hm s ;

V vy vic pht trin t duy hm c ngha quan trng trong dy hc mn

Ton, n va l yu cu ca vic dy hc mn Ton, va l iu kin nng cao

cht lng dy hc mn ton. Vic dy hc cc kin thc mn Ton trnh by theo

t tng hm s c tc dng tt cho vic pht trin t duy hm cho HS. ng thi

c th rn luyn nhiu k nng gii ton v ng dng kin thc ton cho HS trong skt hp pht trin t duy hm.

Tuy nhin, c th ni rng thc t dy hc trng THPT hin nay, ch

BPT v BT vn l mt ch kh i vi HS. Cc em vn thng mc phi mt

s sai lm trong qu trnh gii BPT hay BT . a s cc em cha c k nng thnh

tho gii BPT v BT theo quan im hm. ng thi phng php hm s

c th hin th no cho hp l vn lun gy kh khn cho cc em.


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Nu xy dng c cc bin php s phm v s dng cc bin php

nhm rn luyn k nng gii bi ton bt ng thc v bt phng trnh theo quan

im hm s cho HS THPT trong qu trnh dy hc s gp phn nng cao cht

lng dy hc mn ton v i mi phng php dy hc trong giai on hin nay.7. D kin cu trc ca ti

Ngoi phn M u v Kt lun, ni dung ca ti gm 3 chng:

Chng 1: C S L LUN V THC TIN

1.1. K nng v k nng gii ton cho HS THPT

1.1.1. K nng

1.1.2. K nng gii ton

1.2. Hm s v quan im hm s trong dy hc ton trng THPT

1.2.1. Khi nim hm s

1.2.2. Quan im hm s trong dy hc ton trng ph thng

1.3. Thc tin rn luyn k nng gii bi ton BT v BPT cho HS THPT

theo quan im hm s.


trng ph thng.

1.3.2. Nhng kh khn sai lm thng gp ca HS trong khi gii BPT v

BT theo quan im hm s

1.4. Kt lun chng 1

Chng 2: MT S BIN PHP RN LUYN K NNG GII BI

TON BT NG THC V BT PHNG TRNH CHO HC SINH

THNG THEO QUAN IM HM S

2.1. Tng quan v bt ng thc v bt phng trnh trong chng trnhTon THPT

2.1.1. nh ngha bt ng thc v bt phng trnh

2.1.2. Cc phng php thng thng gii bt phng trnh v chng

minh bt ng thc

2.1.3. Cc k nng cn rn luyn cho HS khi hc ni dung ny.


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Chng 1

C S L LUN V THC TIN

1.1.

K nng v k nng gii ton cho HS THPT1.1.1. K nng

1.1.1.1. Khi nim k nng

C nhiu cch nh ngha khc nhau. Nhng nh ngha ny thng bt

ngun t quan nim chuyn mn v gc nhn c nhn ca mi ngi vit. Tuy

nhin tt c chng ta u tha nhn rng k nng c hnh thnh khi chng

ta p dng kin thc vo thc tin. K nng hc c do qu trnh lp i lp

li mt hoc mt nhm hnh ng nht nh no . K nng lun c ch

ch v nh hng r rng.

Theo Tm l hc i cng: K nng l nng lc s dng cc d kin,

cc tri thc hay kh nng sn c, nng lc vn dng chng pht hin nhng

thuc tnh bn cht ca cc s vt v gii quyt thnh cng nhim v l lun

hay thc hnh xc nh (xem [ 8 ]).

Theo Tm l hc la tui v Tm l hc s phm: K nng l kh

nng vn dng kin thc (khi nim, cch thc, phng php) gii

quyt mt nhim v mi.

T in Ting Vit khng nh: K nng l kh nng vn dng nhng

kin thc thu nhn c trong mt lnh vc no vo thc t (xem [6 ]).

Trong Ton hc th k nng c hiu l kh nng gii cc bi ton,

thc hin cc chng minh cng nh phn tch, c ph phn cc li gii cng

nh chng minh nhn c.

Nh vy, d pht trin di gc no th k nng vn l kh nng ca

ch th thc hin thun thc mt hay mt chui hnh ng trn c s hiu

bit (kin thc hoc kinh nghim) nhm to ra kt qu mong i. Ni n k

nng l ni n cch thc, trnh t thc hin cc thao tc hnh ng t

c mc ch nh. K nng chnh l kin thc trong hnh ng.


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Tuy nhin, trong thc t dy hc, HS thng gp kh khn trong vic

vn dung kin thc (khi nim, cch thc, phng php) gii quyt cc

bi tp c th. HS thng kh tch ra nhng chi tit th yu, khng bn cht ra

khi i tng nhn thc, khng pht hin nhng thuc tnh, mi quan h vn

c gia kin thc v i tng. S d nh vy l do kin thc khng chc

chn, khi nim cng tr nn kh hiu, khng gn lin vi c s ca k nng.

1.1.1.2. S hnh thnh v cc yu t nh hng n s hnh thnh k nng

a)S hnh thnh k nng

Bt c mt k nng no cng c hnh thnh trn c s ca cc

kin thc bit. Chnh v vy, c k nng, trc ht HS cn c cc kinthc c s. K nng c hnh thnh nhanh hay chm l do kh nng nhn

thc v vn kin thc ca ngi hc. K nng ch c hnh thnh thng qua

qu trnh t duy gii quyt nhim v t ra.

Mt k nng s c hnh thnh trong qu trnh thc hin nhiu ln,

lp i lp li mt khi nim, cch thc hay phng php no .Nn hnh

thnh k nng cho HS cn cho HS luyn tp nhiu ln hnh thnh k nng.b) Cc c im c bn ca k nng:

Theo [7], k nng c cc c im c bn sau:

Bt c k nng no cng phi da trn c s l thuyt, l kin thc, bi

v cu trc ca k nng bao gm: hiu mc ch, bit cch thc i n kt qu, hiu

nhng iu kin trin khai cc cch thc .

Kin thc l c s ca cc k nng khi cc kin thc phn nh y

cc thuc tnh bn cht ca i tng, c thc nghim trong thc tin v tn ti

trong thc vi t cch ca hnh ng.

K nng ca con ngi khng phi l yu t bt bin trong sut cuc i m

ph thuc vo ngi hc thng qua hot ng ca h trong mi quan h ca h vi

cng ng.

Gii mt bi ton cn tin hnh mt h thng cc hnh ng c mc ch. Do

ch th gii ton cn phi nm vng tri thc v hnh ng thc hin hnh ng


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theo cc yu cu c th ca tri thc bit hnh ng c kt qu trong nhng iu

kin khc nhau. C th hiu k nng gii ton l kh nng vn dng c mc ch

nhng tri thc v kinh nghim c vo gii nhng bi ton c th, thc hin c kt

qu mt h thng hnh ng gii ton i n li gii mt cch khoa hc. hnh thnh c k nng trc ht cn c kin thc lm c s cho vic

hiu bit, luyn tp tng thao tc ring r cho n khi thc hin c hnh ng

theo ng mc ch yu cu. K nng ch c hnh thnh thng qua qu trnh t

duy gii quyt nhng nhim v t ra. S d dng hay kh khn khi vn dng

kin thc (hnh thnh k nng ) ty thuc vo kh nng nhn dng kiu ton, pht

hin nhn thy trong cc d liu cho ca bi ton, c nhng thuc tnh nhng

quan h l bn cht thc hin gii bi ton cho.V d 1: Gii bt phng trnh.

1x + 2 3x + 50 3x 12 (1)

Nu gii bi ton bng cc phng php thng thng tc l dng

phng php bin i tng ng th bi ton s tr nn phc tp hn.

Ta nhn thy, tng cc bnh phng cc cn thc v tri l mt s khng

i ( 1x )2 + ( 2 3x )2 + ( 50 3x )2 = 48

V tri ca (1) c dng a1b1 + a2b2+ a3b3 Trong BT bunhiacopxki. T

, ta ngh n vic s dng BT bunhiacopxki gii quyt bi ton.

Nu ta xem a1= 1x , a2 = 2 3x , a3 = 50 3x v b1=b2=b3=1 th ta

c:

(1. 1x + 1. 2 3x +1. 50 3x )

2 2 21 1 1 .48

1x + 2 3x + 50 3x 12 lun ng.

Vy tp nghim ca BPT chnh l iu kin cho cc biu thc trong cn

xc nh.

c) Cc yu t nh hng n s hnh thnh k nng

Trong DH Ton, cc nh nghin cu ch ra cc yu t nh hng

n s hnh thnh k nng l (xem [ 8 ]):


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Ni dung bi ton: Nhim v t ra c tru tng ha hay b che

ph bi nhng yu t ph lm lch hng t duy c nh hng n s hnh

thnh k nng.

Tm th v thi quen cng nh hng n s hnh thnh k nng.

Vic to ra tm th thun li cho HS s gip HS d dng trong vic hnh

thnh k nng.

Kh nng khi qut nhn i tng mc cao hay thp.

1.1.2. K nng gii ton

1.1.2.1. Khi nim k nng gii ton

Gii mt bi ton l tin hnh hnh ng c mc ch, v th ch thbi ton cn phi nm vng tri thc v hnh ng, thc hin hnh ng theo

yu cu c th ca tri thc , bit hnh ng c hiu qu trong nhng iu

kin khc nhau. Trong gii ton, theo ti k nng gii ton ca HS l kh

nng vn dng c mc ch nhng tri thc v kinh nghim c vo nhng

bi ton c th, thc hin c kt qu mt h thng hnh ng gii ton i

n li gii bi ton mt cch khoa hc. thc hin tt mn Ton trng THPT, mt trong nhng yu cu

c t ra l: V tri thc, k nng, cn ch nhng tri thc, phng php,

c bit l tri thc c tnh cht thut ton v nhng k nng tng ng. Chng

hn: tri thc v k nng gii bi ton bng cch lp phng trnh, tri thc, k

nng chng minh ton hc, k nng hot ng v t duy hm. .

K nng gii ton l kh nng vn dng cc tri thc ton hc giicc bi tp ton bng suy lun hay chng minh.

Cn ch l ty theo ni dung kin thc ton hc m c nhng yu

cu rn luyn k nng khc nhau.

1.1.2.2. Vn rn luyn k nng gii ton cho HS THPT

Theo Nguyn B Kim ([ 9 ]) mt trong cc mc tiu quan trng ca

vic DH Ton trng ph thng l Trang b tri thc, rn luyn k nng, c


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bit l k nng t duy cho HS. Nh vy, vic truyn t kin thc v rn

luyn k nng gii ton cho HS THPT c coi l nhim v quan trng. Rn

luyn k nng ni chung v k nng ton hc ni ring l mt yu cu quan

trng, n m bo mi lin h gia hc v hnh. Thc tin cho thy, vic dyhc s khng t kt qu cao nu HS ch bit hc thuc lng cc khi nim,

nh ngha, nh l m khng bit p dung gii cc bi ton khc nhau.

Tuy nhin, rn luyn tt k nng gii ton cho HS THPT chng ta

cn trang b cho HS mt khi lng kin thc nht nh, gi l kin thc c

s. Bi l, bt c k nng no cng phi da trn c s ca l thuyt, l

kin thc ti thiu m ngi hc cn c khi hc cc ni dung . Kin thc lc s ca k nng l nhng kin thc phn ng v y cc thuc tnh bn

cht, ng thi c th nghim v p dng thng xuyn, t s hnh

thnh ln k nng.

Dy ton l dy kin thc, k nng, tnh cch cho HS. Vic hnh

thnh v rn luyn cho HS cc k nng gii ton nhm t c nhng yu

cu sau :

Gip HS hnh thnh v nm vng nhng mch kin thc c bn

xuyn sut chng trnh.

Gip HS pht trin nng lc tr tu

Coi trng vic rn luyn kh nng tnh ton trong gi hc. l s

pht trin tr tu cho HS qua mn ton gn lin vi vic rn luyn cc k nng

thc hnh.

Gip HS rn luyn cc phm cht o c v thm m: tnh kin tr,

cn thn, chnh xc, v cc thi quen t kim tra nh gi c t mnh hn

ch cc sai lm.

Trong dy hc Ton cn rn luyn cho HS cc nhm kin thc c

bn sau:

Nhm k nng chung


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Nhm k nng thc hnh

Nhm k nng t duy

K nng suy lun, d on

K nng tng hp: Lin h cc d kin trong bi ton, tm tt ni

dung bi ton, xut nh hng gii quyt bi ton.

K nng phn tch

K nng s dng thng tin

V d2: Gii bt phng trnh

7 3

1

x

x

6 4

1

x

x

(1)

HS thc hin li gii ny nh sau:

iu kin: x 1

(1) 7 31

x

x

(x-1) 6 41

x

x

(x-1)

7x-3 6x -4

x -1

i chiu vi iu kin c nghim ca phng trnh l:

x (-1; 1) (1; )

Thc t HS mt cnh gic khi nhn 2 v ca phng trnh (1) vi

f(x) = x 1 m khng quan tm ti du ca f(x) iu ny nh hng trc tip

n chiu ca BPT dn ti kt qu bi ton sai.

Nh vy vic nm vng cc nh l v bin i BPT l quan trng v

cn thit . Cn a ra nhng bi tp HS c th vn dng cc php bin itng ng ny thnh tho.

Lm r s ging nhau v khc nhau gia php bin i tng ng

BPT bi php bin i tng ng trong phng trnh v trnh sai lm khi

p dng.

Tht v ngha nu yu cu HS hc thuc lng cc nh l v cc php

bin i tng ng hoc cch p dng n.


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C th rn luyn k nng gii ton cho HS theo cc con ng khc

nhau

Con ng th nht: Sau khi cung cp v truyn th cho HS vn

kin thc cn thit th yu cu HS p dng gi cc bi ton khc nhau theo

mc tng dn.

Con ng th hai: Dy nhng du hiu c trng, t a ra cc

dng ton v nh hng cc tao tc cn thit gii dng ton .

Con ng th ba: Dy hc cc hot ng tm l cn thit i vi

vic vn dng tri thc.

V d 3: Gii bt phng trnh

2x 2x 3 - 2x 6x 11 > 3 x - 1x (1)

Vi bi ton ny, HS c th d dng tm c iu kin ca BPT

(1x 3)

V vit li BPT di dng :

2x 2x 3 + 1x > 2x 6x 11 + 3 x

n y, HS c hai hng gii quyt,

Hng 1 : Nhn thy y l phng trnh v t c hai v khng m nn

kh du cn ta bnh phng hai v, hng lm ny c th ra p s nhng

chc chn kh phc tp v d nhm ln trong tnh ton.

Hng 2 : Nhn xt cc s hng tng v ca BPT. T , t BPT

di gc nhn theo quan im hm. Tc l, c th s dng cng c hm s d

gii quyt BPT ny khng ?.

(1) 2

1 2x + 1x > 2

3 2x + 3 x (2)

Xt hm s: y= f(t) = 2 2t + ttrn on [0 ; 2 ] ( V x [1; 3 ] )

f(t) =2 2

t

t +

1

2 t > 0 vi mi t [0 ; 2 ]

=> f ng bin trn [0 ; 2 ]. Do t (2) ta c:


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13

f(x-1) >f(3-x) x-1 > 3-x x>2

Kt hp vi iu kin 1 x 3 bt phng trnh c nghim 2


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+ Nhn nhn bi ton di nhiu kha cnh khc nhau, t so snh

cc cch gii vi nhau hiu su sc v vn dng hp l kin thc.

+ Quan st t m v ch tm ra c im bi ton.

+ Tch cc suy ngh, tm ti cch gii ngn gn trong khi gii ton.

Thc t, trong hc tp HS khng ch gp nhng bi ton n gin, tun theo

phng php v cc bc lm r rng m cn gp kh nhiu bi ton phc

tp, khng c phng php sn. i hi HS phi suy ngh tm ra cch gii

ngn gn, cht ch v c o.

V vy, song song vi vic truyn th tri thc ton hc th vic rn

luyn cc k nng cng ng mt vai tr ht sc quan trng, n gp phn bidng t duy Ton hc cho HS.

1.2.

Hm s v quan im hm s trong dy hc ton trng THPT

1.2.1. Khi nim hm s

Khi nim hm s c nhc n chng trnh mn Ton cp

THCS, v n chng trnh cp THPT th vn ny vn tip tc c

nghin cu. lp 10 HS vn tip tc c nghin cu v hm s. y

SGK gii thiu li khi nim hm s 1 cch chnh xc hn, c cp n tp

xc nh ca hm s, ng thi a ra cc khi nim hm s ng bin,

nghch bin, hm s chn, l, v gii thiu thm 1 phng php nghin cu

hm s l kho st s bin thin v v th ca chng. SGK trnh by y

v 2 hm s y =ax + b v y = ax2+bx+c, ngoi ra SGK cn gii thiu thm

v hm s y = ax + b| .

Khi nim hm s c gii thiu chng trnh ton lp 7, v vy

y

SGK lp 10 khng xut pht t v d m gii thiu ngay nh ngha v

cho v d minh ha.

nh ngha trang 32- SGK i s 10 Gi s c 2 i lng bin

thin x v y, trong x nhn gi tr thuc tp D. Nu vi mi gi tr ca x


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15

thuc tp D c mt v ch mt gi tr tng ng ca y thuc tp s thc R th

ta c mt hm s. Ta gi x l bin s, y l hm s ca x. Tp D c gi l

tp xc nh ca hm s.

nh ngha trang 35- SGK i s 10 nng cao : Cho tp hp khc

rng l tp con ca R. Hm s f xc nh trn D l mt quy tc tng ng mi s

x thuc D vi 1 v ch 1 s, k hiu l f(x), s f(x) c gi l gi tr ca hm s

f ti x. Tp hp D gi l tp xc nh, x l bin s hay i s ca hm s.

Sau khi trnh by nh ngha, SGK a ra cc v d v hm s trong

thc t l cc hm s cho bng bng. V cch cho hm s, SGK i s 10

trnh by 3 cch cho hm s : hm s cho bng bng, hm s cho bng biu

, hm s cho bng cng thc.

V d 4 :Khi hc v phn hm s lp 10, gio vin c th trnh by 1

trong s nhng cch cho hm s l hm s cho bng cng thc, v d nh cho

hm s s = 50t, trong vn tc trung bnh ca xe my l 50 km /h, t l thi

gian xe i c trong qung ng s. T hm s trn ta thy c mi quan

h gia cc i lng s v t.

1.2.2. Quan im hm s trong dy hc ton trng ph thngDo c tnh biu th quan h ph thuc ln nhau gia cc i lng,

mt quan h ph bin phn nh bn cht ca hu nh mi hin tng trong

khoa hc cng nh trong cuc sng, hm s khng ch xut hin trong ton

hc m cn dung gii quyt cc vn thc tin. Trong cc gio trnh,

sch gio khoa ton, hm s xut hin trc ht vi t cch l i tng

nghin cu, sau l cng c gii quyt nhiu bi ton thuc nhng ni

dung ton hc khc nh: BT v BPT Cng v vai tr quan trng ca n

m hm s l mt ch xuyn sut cc chng trnh ton bc trung hc ca

nhiu thp k qua. Chng ta bit, trong mn Ton THPT, mt hm s c th

c biu th bng nhng h thng biu t khc nhau. Tuy nhin s dng

mt biu thc Ton hc (hm s cho bi cng thc) v th l hai trong

nhng h thng biu t ph bin nht. Khi bit biu thc xc nh hm s,

ta c th dng cng c ca gii tch i s nghin cu cc tnh cht v phc


22/68

16

tho th ca n. Ngc li, nhn vo th ta c th c c cc tnh cht

ca th nh : chiu bin thin trong tng khong, tnh chn l, tnh tun

hon, tnh b chn, gi tr cc i, cc tiu

1.3. Thc tin rn luyn k nng gii bi ton BT v BPT cho HSTHPT theo quan im hm s.


trng ph thng.

Do s tit hc ch BT v BPT trn lp cn t nhng khi

lng tri thc cn truyn t nhiu ng thi phi ng lch phn phi

chng trnh theo quy nh nn vic m rng, khai thc, ng dng sng to

cc kin thc hc cha c trit su sc. iu ny nh hng n vic

huy ng vn kin thc ca HS, hn ch vic rn luyn tnh tch cc, c lp,

sng to, ca HS trong hc tp, cng nh gio vin khng c nhiu thi gian

gip HS.

Trong thc t, cch dy ph bin hin nay l gio vin vi t cch l

ngi ngi iu khin a ra kin thc ri gii thch chng minh, sau a

ra mt s bi tp ng dng, lm cho HS c gng tip thu vn dng.

Cht lng i tr ca HS cn yu. Nhiu HS b mt cn bn t cp

THCS dn n tnh trng cc em nhn thy ch ny l mt ch rt kh

hiu, tru tng nn khng c hng th trong vic hc v khng c phng

php hc ph hp cho bn thn. Cc em hu nh khng lm bi tp v nh,

li suy ngh, li tnh ton, cha tch cc t duy hot ng tr no tm ti

pht hin vn v gii quyt vn .

Trong gi hc, HS cha ch nghe ging, khng tch cc pht biu,

xy dng bi, ch yu nghe ging, tip thu kin thc mt cch th ng nn

d qun, khng hiu c bn cht vn .


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17

Tuy nhin, trong dy hc ton trng ph thng. Vic pht trin t

duy hm cho HS khng c ngha l thy gio ln lp v ging 1 bi ging v

t duy hm. Nhim v t duy hm khng tn ti c lp so vi cc nhim v

truyn th kin thc khc. Mun pht trin kh nng t duy hm cho HS

thy gio phi thng qua ni dung kin thc quy nh. Trn c s tm

ra gii php pht trin t duy hm cho HS. C th ni, pht trin t duy hm

l mc ch kp.

Thc tin gio dc t duy hm cho HS ph thng hin nay cn gp

kh nhiu nhng kh khn nh : trnh HS cn hn ch v khng ng u,

khi lng kin thc nhiu trong khi s tit hc dnh cho mn ton li khng

nhiu. Hn th, nhng tri thc v hot ng t duy hm khng c quy nh

r rng trong chng trnh nn khng c ging dy mt cch tng minh.

M trong ton hc, vic xem xt cc i tng ton hc mt cch c lp trong

trng thi ri rc s khng thy ht c nhng mi lin h ph thuc hay

mi quan h nhn qu lm cho HS cn lng tng trong vic gii quyt cc bi

ton. Bn cnh , cc ti liu vit v vn ny ni chung cn hn ch, kh

tip cn. V vy cn gy ra khng t nhng kh khn cho c gio vin v HS.

1.3.2. Nhng kh khn sai lm thng gp ca HS trong khi gii BPT v

BT theo quan im hm s

1.3.2.1. Kh khn

Trong chng trnh ton trung hc ph thng, cc bi ton v phn

BT v BPT l mt ch kh, n gy ra nhiu tr ngi i vi cc em HS

trong vic chim lnh tri thc. iu ny cng dn n vic gii cc bi tp ca

HS rt kh khn, cc em cn t ra lng tng, cha c rn luyn v k nng

gii ton, cha kch thch c s ham m tm ti khm ph ca HS.

T HS tip thu kin thc mt cch hnh thc v hi ht qu trnh

bi dng kin thc ton hc theo hng nng cao ca ch BT v BPT


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cho HS cha c lin mch v cha c h thng. Chnh iu lm cho HS

d ht hng v kin thc, vic khai thc mt bi ton cn gp nhiu kh khn,

vic dy hc ca gio vin ch yu da vo kinh nghim ca bn thn.

Mt s kh khn ch yu l :

HS khng bit cch phn chia cc trng hp gii quyt bi ton.

HS khng bit phi xut pht t c s no phn chia cc trng

hp ring thch hp cho vic gii quyt bi ton.

HS khng bit nhn nhn cc bi ton bt phng trnh, bt ng

thc trong mi lin h vi cc bi ton hm s.

1.3.2.2. Mt s sai lm ca hc sinh trong gii ton BT v BPT theo quan im hm

Ni dung BT v BPT l mt ni dung quan trng trong chng trnh

THPT. Tuy nhin, c GV v HS u cha tht s ch trng n vn ny.

V vy, a s HS ch lm cc bi tp mt cch my mc m cha hiu c

bn cht ca vn dn n mc phi mt s sai lm ng tic trong khi gii

cc bi ton bt phng trnh cng nh chng minh BT.

Sau y l mt vi v d in hnh.

V d 5: Gii BPT sau: x2 x 4 + 24 x 2

22 4

x

x (1)

Sai lm thng gp : BPT ( 1)x2 x 4 + 24 x

2 2

2

4 (2 )x x

x

2 2 20

x 4 4 2 4

x

x x x

20 0

2 36 0

x x

xx x

Nguyn nhn sai lm : Php bin i 2 2 2x 4 4 2 4x x x thnh

2 6 0x x l khng tng ng.

Li gii ng : iu kin xc nh ( x 0, -2< x < 2).

Bpt (1) x2 x 4 + 24 x 2 2

2

4 (2 )x x

x


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19

2 2 2

0

x 4 4 2 4

x

x x x

2

2

0

4 0

6 0

x

x

x x

0

2 2

2 3

x

x

x

02 2

xx

Kt lun : f(x) g(x) f(x) + h(x) g(x) + h(x) (1)

h(x) D vi D l tp xc nh ca f(x) g(x).

f(x) + h(x) g(x) + h(x) f(x)g(x), vi x thuc tp xc nh ca (1).

V d 6: CMR:

2 2 2 2 2

( )a b c d e a b c d e vi mi sthc , , , ,a b c d e.

Li gii:Theo C-si ta c:2 2 2 2

2 2 2 2, , ,4 4 4 4

a a a ab ab c ac d ad e ae

Cng cc BT trn ta ciu phi chng minh.

nh gi:

y HS vn dng bt ng thc C-si l sai, v cc s c th m.

Tuy nhin, mi BT trn u ng nhng khng phi theo C-si, m do

2( ) 0,...2

ab

V d 7: Tm GTLN ca biu thc (2 )(2 )M x a x b x vi a, b

dng, phn bit v0 < x < 2a, 0 < x < 2b

Li gii:

V 1

.2 (2 )(2 )2

M x a x b x

p dng BT Csi cho cc s2x, 2a-x, 2b-xnn M ln nht khi chng

bng nhau, nhng iu khng xy ra nn M khng c GTLN .


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nh gi: iu ny sai logic v khi 3 s bng nhau th c gi tr ln

nht, cn khi khng bng nhau th cha kt lun c g.

V d 8:Tm m sao cho

2

2

2 3 2 12 2

x mx mx mx

xR (*)

(*)x2 2mx + 3m + 2 2 x2 m x 2 x R

x2+ mx 3m 0 0 m2+ 12m 0 -12 m 0

Sai lm khi nhn 2 v ca (*) vi 2 x2 m x 2 khi cha bit du ca

biu thc ny.

nh gi chung : Sai lm do khng ch gi thit ca nh l suy ravi vng p dng, lm dng suy din nhng mnh khng ng hoc xt

thiu trng hp cn bin lun.

Tm li nhng sai lm ca hc sinh cn c thy c ch ra hoc hng dn

nhn ra t sa cha sai lm ca mnh. Thy c cn c bin php khi ging dy

nh nhn mnh kin thc, ra nhiu bi tp nhiu dng khc nhau. Cn c s kt

hp nhiu phng php v nhiu v d minh ha cho cng thc mt cch hp l. S

dng cc cng c dy hc c hiu qu.

1.4. Kt lun chng 1

Trong chng 1, ti trnh by mt s cch hiu bit v khi nim

k nng, k nng ton hc, hm s v quan im v hm s. Vic rn luyn

k nng gii bi ton bt ng thc v bt phng trnh cho HS l rt cn

thit bi qua gip HS hc tp tch cc, kch thch tnh sng to ca HS

trong hc tp v trong cuc sng.


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Chng 2

MT S BIN PHP RN LUYN K NNG GII BI TON

BT NG THC V BT PHNG TRNH CHO HC SINH

THNG THEO QUAN IM HM S

2.1. Tng quan v bt ng thc v bt phng trnh trong chng trnh

Ton THPT

2.1.1. nh ngha bt ng thc v bt phng trnh

2.1.1.1. Bt ng thc

Mt bt ng thc l mt pht biu v quan h th t gia hai i

tng

K hiu a b c ngha l anh hnbv

K hiua b c ngha l aln hnb.

Nhng quan h ni trn c gi l bt ng thc nghim ngt; ngoi

ra ta cn c a b c ngha l a nh hn hoc bngbv a b c ngha

l aln hn hoc bngb.

Ngi ta cn dng mt k hiu khc ch ra rng mt i lng lnhn rt nhiu so vi mt i lng khc.

K hiu a >> b c ngha l aln hnbrt nhiu.

Cc k hiu a, b hai v ca mt BT c th l cc biu thc ca cc

bin. Sau y ta ch xt cc BT vi cc bin nhn gi tr trn tp s thc

hoc cc tp con ca n.

Nu mt BT ng vi mi gi tr ca tt c cc bin c mt trong

BT , th bt ng thc ny c gi l BT tuyt ihay khng iu kin.Nu mt bt ng thc ch ng vi mt s gi tr no ca cc bin, vi

cc gi tr khc th n b i chiu hay khng cn ng na th n c gi l

mt BT c iu kin. Mt BT ng vn cn ng nu c hai v ca n

c thm vo hoc bt i cng mt gi tr, hay nu c hai v ca n c

nhn hay chia vi cng mt s dng. Mt BT s b o chiu nu c hai v

ca n c nhn hay chia bi mt s m.


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2.1.1.2. Bt phng trnh

Bt phng trnh mt n trn trng s thc c nh ngha l:

mt mnh cha bin x so snh hai hm s f(x) v g(x) trn trng s thc

v c vit di mt trong cc dng sau

f(x) > g(x) , f(x) < g(x) , f(x) g(x) , f(x) g(x)

Giao ca hai tp xc nh ca hai hm s f(x) v g(x) l tp xc nh

ca bt phng trnh.

Cc bt phng trnh trn u c th chuyn v cc bt phng trnh

dng trng ng nh f(x) > 0 (hoc f(x) 0)

Bin x trong bt phng trnh c gi l n (hay i lng cn tm).

Gii bt phng trnh l ta i tm tp nghim ca bt phng trnh

. y, ta xt bt phng trnh dng tng qut f(x) > 0

Nu vi gi tr x = a, f(a) > 0 l mt bt phng trnh ng th ta ni

a nghim ng ca bt phng trnh f(x) > 0 hay a gi l nghim ca bt

phng trnh .

Tp hp tt c cc nghim ca bt phng trnh c gi l tp

nghim ca bt phng trnh . Trong mt s ti liu ngi ta cn gi tp

nghim ca bt phng trnh l nghim ca n.

Bt phng trnh nhiu n: nh ngha bt phng trnh nhiu n c

th c suy ra t m rng ca bt phng trnh mt n vi n bin trn

hoc trn tp bt k ca bin x nhng cc hm f(x) v g(x) phi nhn cc gi

tr trn tp sp th t ton phn. Vic gii BPT hay mt h BPT l qu trnh bin i cc BPT

thnh nhng BPT dng tng ng m n gin hn nhng BPT ban u.

Tuy nhin, qu trnh ny thng phc tp hn i vi cc bt phng trnh v

t hay bt phng trnh cha du gi tr tuyt i, BPT logarit, lng gic,

trch nhm ln, khi gii cc bt phng trnh HS cn xc nh

mt cch chnh xc v y cc iu kin xc nh bt phng trnh


29/68

23

nhn c lun tng ng vi bt phng trnh cho. Khi gp cc iu

kin vi bt phng trnh mi ta nhn c mt h bt phng trnh mi

tng ng vi bt phng trnh cho. T chng ta c th a ra kt

lun y nht v tp nghim ca bt phng trnh.2.1.2. Cc phng php thng thng gii bt phng trnh v chng

minh bt ng thc

Vic xem xt, nghin cu cc bi ton trong ton hc s cp bng

cch ghp thnh tng lp bi ton c th gii c bng cng mt phng

php l mt vic lm cn thit v c ngha. Trn c s l thuyt v bi tp

trong sch gio khoa cng nh mt s sch tham kho khc. C th lit k

mt s phng php gii bt phng trnh nh sau:

Phng php bin i tng ng.

Phng php t n ph.

Phng php hm s.

Phng php th.

Phng php xt iu kin cn v .

Phng php nh gi....

V mt s cch chng minh bt ng thc nh:

Phng php dng nh ngha.

Phng php dng cc php bin i tng ng.

Phng php dng cc bt ng thc ph.

Phng php vn dng cc bt ng thc c bn.

Phng php s dng tnh cht bc cu.

Phng php s dng tnh cht ca t s.

Phng php dng hnh hc v th.

Phng php i bin s.

Phng php quy np


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24

V d 9:. Cho , ,x y z l ba s dng v 1x y z . Chng minh

rng: 2 2 22 2 2

1 1 182x y z

x y z

+ Hy vo v tri ca bt ng thc, tng bnh phng ca i

lng 22

1x

x gi cho cc bn ngh ti bt ng thc no? Du ng thc xy

ra khi no?

Bt ng thc Bunhiacopxki cho hai b s ;a b v ;c d :

22 2 2 2a b c d ac bd

ng thc xy ra khi:a b

c d hay hai b ;a b v ;c d t l.

+ Vai tr ca , ,x y z nh nhau nn p dng u cho , ,x y z.

+ ng thc xy ra khi no? (khi1

3x y z ), hy tm b s t l vi

b s 1;xx

? 13;3

hoc 1;9 ;...

+ Cn p dng bt ng thc Bunhiacopxki cho 1;xx

v b s no?

p dng bt ng thc Bunhiacopxki cho hai b s 1;xx

v 1;9 , ta c:

2 2

2 2 22

1 91 9 x x

x x

hay 22

1 982 x x

x x

+ Vn dng tng t vi 22

1y

y v

22

1z

z v suy ra bt ng thc no?

2 2 22 2 2

1 1 1 1 1 182 9x y z x y z

x y z x y z

+ Khi cn chng minh bt ng thc trung gian no?


31/68

25

1 1 19 82x y z

x y z

+ Cc biu thc 1 1 1

x y z v x y z gi cho ta ngh n bt ng

thc quen thuc no?

1 1 19 9x y z

x y z

hay1 1 1 9

x y z x y z

+ Vn dng vo bi ton ta cn chng minh bt ng thc no?

1 1 1 8182

x y z x y z

+ Tch 81

x y zx y z

l hng s gi cho cc bn ngh ti bt ng

thc no?

Cauchy cho hai s dng 81

2 81x y zx y z

nhng khng suy

ra bt ng thc cn chng minh.

+ Hy xem li du ca bt ng thc xy ra? (khi 1

3x y z hay

1x y z ) khi cn p dng bt ng thc Cauchy cho hai s no?

1 1 1 12

x y z x y z

Ta cn chng minh 80 80

x y z

hay

1x y z (theo gi thit), t suy ra bt ng thc cn chng minh.

* Nhn xt: Qua li gii trn cho thy nu bit nhn nhn vn theo

nhiu hng, ngi gii ton bit lin tng, huy ng kin thc ph hp s

mang li mt cch gii quyt vn tt p nht.

Nh vy bit nhn bi ton di nhiu gc , kha cnh khc nhau

khng nhng cng c c kin thc m cn rn luyn, bi dng thm kh

nng huy ng kin thc HS.


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V d 10 : Cho hai s thc x v y tha mn: 2x y .

Chng minh rng: 2 2 2x y

p dng bt ng thc Bunhiacopxki, ta c:

22 2 2 2 2 2 2 21 1 2 4 2x y x y x y x y .

Gi m l mt gi tr ca biu thc 2 2x y . Khi h phng trnh sau c

nghim:

2 2

2x y

x y m

(1)

Ta c, h (1)

2

22 24

4 222

x yx y x ym

xy m xyx y xy m

Do h (1) c nghim khi v ch khi4

4 4. 22

mm

Vy 2 2 2x y .

Trong mt phng Oxy xt ng thng c phng trnh:2 0x y v ;M x y . Khi : M v 2 2OM x y

Ta c: 0 0 2

; 21 1

d O

v ;OM d O (v M), suy ra

2 2 2 22 2x y x y .

C: 2 2 21 2 2 1 2 2x x x x x x

Tng t: 2 1 2y y

T suy ra: 2 2 2 22 2 2x y x y x y .

T gi thit 2x y ta suy ra 2y x

Do : 2 22 2 22 2 0 2 1 0x y x x x , ng.


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Nhn xt: Qua li gii trn cho thy nu bit nhn nhn vn theo

nhiu hng, ngi gii ton bit lin tng, huy ng kin thc ph hp s

mang li mt cch gii quyt vn tt p nht.

Nh vy bit nhn bi ton di nhiu gc , kha cnh khc nhau

khng nhng cng c c kin thc m cn rn luyn, bi dng thm kh

nng huy ng kin thc HS.

Khi ng trc mt bi ton gii bt phng trnh hay chng minh bt

ng thc bt k no th vic nh hng phng php gii bi ton

ng vai tr quyt nh thc hin li gii bi ton . Tuy nhin, vic nhhng phng php gii cc dng ton ny rt a dng, c nhng bi ton c

th gii bng nhiu phng php khc nhau. Vn ch l la chn phng

php no l ti u nht trnh by li gii cho bi ton mt cch ngn gn

v chnh xc nht.

2.1.3. Cc k nng cn rn luyn cho HS khi hc ni dung ny.

Chng ta u bit, ch bt phng trnh v bt ng thc l mt ch

kh trong chng trnh ton ph thng. Chnh v vy, mi HS cn c trang

b cho mnh nhng k nng tt nht c th hc tt ni dung ny.

Nhng sai lm m HS thng mc khi gii cc bt phng trnh l

khi HS vi phm cc qui tc bin i phng trnh, bt phng trnh tng

ng. t tha hay thiu nhng iu kin ca bt phng trnh u c th dn

n nhng sai lm. Hay c khi, sai lm c th ch HS chn cha ng cch

gii quyt bi ton

i vi vic chng minh cc bt ng thc, sai lm thng bt

ngun t vic vn dng cc bt ng thc c in m khng ch n cc

iu kin bt ng thc ng. Hoc s dng sai st cc quy tc suy lun

khi t bt ng thc ny suy ra bt ng thc kia.


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C nhiu kiu phn chia k nng ph hp vi tng mng kin thc,

tng ni dung mn hc. Tuy nhin, tu trung li HS cn c rn luyn

nhng k nng c bn nh k nng nhn bit, k nng nhc li, k nng d

on, k nng phn loi v k nng thc hnh vn dng, y l nhng k

nng v cng quan trng, n khng ch cn thit cho vic hc tp tt mn

ton m cn cn thit trong sut chng trnh ph thng, tt c cc ni dung

v tt c cc mn hc.

S phn chia cc k nng nh trn ch mang tnh cht tng i.

Thc t, trong dy hc, ta khng ch dy cho HS tng k nng n l trn

m cn phi rn luyn k nng phc hp. Tc l, tng ni dung, tng mng

kin thc c th ta khng ch rn luyn tng mng k nng c bn n l v

mt k nng c th l hn hp ca nhiu k nng c bn. Chng hn nh, k

nng v th bao gm c k nng nhn thc, k nng phn loi v k nng

thc hnh. V v c mt th hm s hon chnh HS cn phi nhn bit

c l loi hm s no ( k nng phn loi ), V th hm s nh th

no (k nng nhn bit ), V cc bc v nn th ( K nng thc hnh ).

Ngoi ra, cng rn c cho HS tnh cn thn, kin tr v nn th hm

s p v chnh xc.

i vi ti bt phng trnh v bt ng thc, cn ch trng rn

luyn cho HS cc k nng thuc nhm k nng nhn thc v k nng vn

dng. C th k ra mt s k nng nh :

K nng tnh ton.

K nng thc hin cc php bin i.

K nng vn dng cc phng trnh mu v cc bt ng thc c bn.

K nng d on.

K nng s dng th.


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29

2.2. Mt s bin php rn luyn k nng gii bi ton bt ng thc theo

quan im hm s.

2.2.1. Rn luyn k nng vn dng o hm chng minh bt ng thc

2.2.1.1. Cc kin thc c bn cn thit.

1. nh l 1: Cho hm s y = f(x) xc nh v lin tc trn [a; b].

*) Nu ( ) 0, ;f x x a b th f(x) ng bin trn [a; b] v khi ta c

; ;min ( ) ( ); m ax ( ) ( )

x a b x a bf x f a f x f b

*) Nu ( ) 0, ;f x x a b th f(x) nghch bin trn [a; b] v khi ta c

; ;min ( ) ( ); m ax ( ) ( )

x a b x a bf x f b f x f a

2. nh l 2: (nh l Fermart)

Gi s hm s y = f(x) xc nh trn mt ln cn b ca 0 ;x a b v

c o hm ti im 0x . Khi nu hm s y = f(x) t cc tr ti 0x th

0( ) 0f x .

3. nh l 3: (iu kin hm s c cc tr)

Cho hm s y = f(x) xc nh trn [a; b] v 0x . Trong mt ln cn

b ca 0x , nu 0( )f x thay i du khi x qua 0x (c th khng tn ti 0( )f x )

th f(x) t cc tr ti 0x .

*) Nu 0 0( ) 0, ;f x x x x v 0 0( ) 0, ;f x x x x th 0x l im

cc tiu.

*) Nu 0 0( ) 0, ;f x x x x v 0 0( ) 0, ;f x x x x th 0x l im

cc i.


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4. nh l 4: Gi s y = f(x) xc nh trn [a; b] v 0 ;x a b . Trong

mt ln cn b ca 0x , hm s y = f(x) c o hm cp hai lin tc,

ng thi 0( ) 0f x

v ( ) 0f x th 0x

l mt im cc tr ca hm s.

*) Nu 0( ) 0f x v ( ) 0f x th 0x l mt im cc tiu ca hm s.

*) Nu 0( ) 0f x v ( ) 0f x th 0x l mt im cc i ca hm s.

2.2.2.2. ng dng o hm chng minh cc bi ton bt ng thc.

A. Bt ng thc mt bin s

1, Dng 1: Kho st trc tip cc tr ca hm s tm tp gi tr

ca hm s

V d 11:( Thi HSG Quc gia, 1992)

Chng minh rng vi mi s t nhin n > 1 ta c

1 1 2n n

n nn n

n n .

Gii: t *0;1 ,n n

x n Nn

. BT cn chng minh tr thnh

1 1 2, 0;1n nx x x .

Xt hm s ( ) 1 1n nf x x x lin tc 0;1x c

1 1 1

'( ) 0, 0;11 1

(1 ) (1 )

f x xn n nn n

x x

Vy f(x) nghch bin [0; 1) nn f(x) < f(0) = 2, 0;1x (pcm).

V d 12: (H An ninh, 1997)

Cho n l s l ln hn 3. Chng minh rng vi mi 0x ta c

2 2 31 ... 1 ... 1

2! ! 2! 3! !

n nx x x x xx x

n n

.

Gii: t


37/68


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32

1. 1

2 .nx x

ne vi mi 0;1x .

Gii: Ta c

21 1. 1 (1 ) (*)22 .

n nx x x xnene

Xt hm s 2( ) (1 )nf x x x vi 0;1x . Ta c

2 1( ) 2 (2 1)nf x x n n x

Nn ta c bng bin thin

x0

22 1

nn

1

f(x) + 0 -

f(x)

Vy2

2 1(0;1)

(2 )ax ( )

(2 1)

n

n

nm g x

n

.

Ta chng minh

2 12

2 1

(2 ) 1 2 1

(2 1) 2 2 1

nn

n

n n

n ne n e

2 1

2 1(2 1) ln(2 1) ln(2 ) 1

2

nn

e n n nn

1ln(2 1) ln(2 ) (2)

2 1n n

n


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33

p dng nh l Lagrange cho hm s ( ) lnf x x trn [2n; 2n+1] suy ra

tn ti 2 ;2 1c n n thuc sao cho(2 1) (2 )

( )2 1 2

f n f nf x

n n

. Suy ra

1 1ln(2 1) ln(2 ) (3)

2 1n n

e n

.

T (1), (2) v (3) suy ra pcm.

2, Dng 2: S dng tnh n iu

Trong mt s bi ton c th phi o hm nhiu ln lin tip thm ch

phi kho st thm hm s ph. Ta thng s dngf(x)ng bin trn [a; b] thf(x) > f(a)vi mi x > a.

f(x)nghch bin trn [a; b] thf(x) > f(b)vi mi x < b.

V d 14: Cho a, b, x > 0 v a b . Chng minh rng

b x ba x a

b x b

Gii:

Xt hm s ( ) , 0.b x

a xf x x

b x

Khi ln ( ) lna x

f x b xb x

.

Suy ra: ln ( ) ( ) lna x

f x b xb x

( )ln ( ) . ln

( )

f x a x b x a x a x b ab x

f x b x a x b x b x a x


40/68

34

( ) ( ) ln

ln . ( )b x b x

a x b af x f x

b x a x

a x a x b a a xg x

b x b x a x b x

trong ( ) lna x b a

g xb x a x

. Ta c

2

2 2 2

( )( ) . 0.

( ) ( ) ( ) ( )

b x b a a b a bg x

a x b x a x a x b x

Do g(x)nghch bin trn 0; . Suy ra

( ) lim ( ) lim ln 0x x

a x b ag x g x

b x a x

.

Vy ( ) 0, 0f x x , nn f(x) ng bin trn 0; . Suy ra f(x) >

f(0)(pcm).

3, Dng 3: Kt hp vi cc BT khc nh BT AM-GM, BT

Cauchy-Schwarz, BT Chebyshes,

V d 15: Chng minh rng nu 02

x

th

s inx t anx 12 2 2x .

Gii:

p dng BT AM-GM ta csinx t anx s inx t anx

2 2 2. 2 .2 . Ta chngminh

s inx t anx 1 s inx t anx 22 2 .2 2 2 2 s inx t anx 2x x x .

Xt hm s ( ) sinx tanx 2f x x lin tc trn 0;2

, c


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22 2

1 1( ) cos 2 cos 2 0, 0;

os os 2f x x x x

c x c x

.

(v vi 0;2

x

th2cos osx c x v theo BT AM-GM ta c

22

1cos 2

osx

c x )

Do f(x) ng bin trn 0;2

. Suy ra ( ) (0) 0f x f , hay

s inx t anx 2x vi mi 0;

2

x

(pcm).

V d 16: ( Olympic 30 - 4 - 1999)

Chng minh rng

3sinx

cos , 0;2

x xx

.

Gii: Ta bin i3 3

2 33

sinx sincos cos sin .tan - x 0 (1)

xx x x x

x x

.

Xt hm s 2 3( ) sin . tan -f x x x x vi 0;2

x

. Ta c

2 2 22sin tan 3f x x x x .

p dng BT 22 2 23( )a b c a b c ta c

22 2 2 2 21sin sin tan 3 2sin tan 3

3f x x x x x x x x .

t ( ) 2sin tan 3 , 0;2

g x x x x x

, th


42/68

36

232 2 21 1 1

2 cos 3 cos cos 3 3 cos . 3 0cos cos cos

g x x x x xx x x

vi mi 0;2

x

, nn g(x) ng bin trn 0;2

. Suy ra

0 0, 0;2

g x g x

.

Do 2 21 3 3 0

3f x x x nn f(x) ng bin trn 0;

2

. Suy ra

0 0, 0; 2f x f x

.

Nhn xt: Khi trong BT c cha cc loi hm s khc nhau ta

thng c lp mi loi hm s d xt du ca o hm, hoc ta c th o

hm lin tip kh bt mt loi hm s nh trong bi ton 5.

B.

Bt ng thc c hai hay nhiu bin s

chng minh BT c cha nhiu bin s bng phng php o hmth iu quan trng nht l chng ta a c v mt bin v kho st hm s

theo bin .

1. Dng 1: Kho st hm c trng

V d 17: Chng minh rng

a)2

12,

1

xx

x x

b) 2 2 21 1 1 3, , ,x x y y z z x y z

tha mn x + y + z = 3.

Gii: a) Xt hm s 2

1,

1

xf x x R

x x

. Ta c

2 2

3 1

2 1 1

xf x

x x x x


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37

v

0 1; lim 1, lim 1x x

f x x f x f x

Ta c bng bin thin

x 1

f(x) + 0 -

f(x) 2

-1 1

T bng bin thin suy ra 1 2,f x f x .

b) p dng cu a ta c

22

1 12, 1 1 (1)

21

xx x x x

x x

Tng t ta c

2

2

11 1 2

21

1 1 32

y y y

z z z

Cng tng v cc BT (1), (2) v (3) ta c

2 2 2 1

1 1 1 3 32x x y y z z x y z (pcm).

Trong mt s bi ton ta c th nhn thy ngay hm c trng, tuy

nhin mt s bi ta cn phi bin i mi nhn thy hm c trng. Xt v d

sau

V d 18: Cho A, B, C l ba gc ca mt tam gic nhn. Chng minh rng

tan tan tan 6 sin sin sin 12 3A B C A B C .


44/68

38

Gii: Xt hm s

tan 6sin 7f x x x x vi 0;2

.

Ta c

2 2

cos 1 3cos 1 2cos 116 cos 7

os os

x x xf x x

c x c x

.

V 0;2

x

nn

0 2cos 1 03

f x x x

.

Lp bng bin thin ca f(x) trn 0;2

ta c

0;

2

7min 4 3

3 3f x f

.

p dng vo bi ton ta c

73 4 3 3

tan tan tan 6 sin sin sin 12 3.

f A f B f C

A B C A B C

Nhn xt: Trong BT trn A, B, C bnh ng nn ta d dng kim tra

c du bng xy ra khi v ch khi3

A B C

. V vy ta cn chn mt

hm s c dng

tan 6sinf x x x kx

sao cho 03

f

. Do k = -7 v ta tm c hm c trng cn xt.

V d 19: (H Quc gia H Ni, 2000)

Cho a + b + c = 0. Chng minh rng

8 8 8 2 2 2a b c a b c .


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39

Gii: Xt hm s 3

2 2 2 ln 2x xf x x trn R. Ta c

2

3. 2 .ln 2 2 .ln 2 2ln 2 2 1 3.2 2 ln 2x x x xf x

v

0 2 1 0xf x x .

Ta c bng bin thin

x 0

f(x) - 0 +

f(x)

0

Suy ra 0, 0f x x R f a f b f c

8 8 8 (2 2 2 ) 2 ln 2 0

8 8 8 2 2 2

a b c a b c

a b c a b c

a b c

Nhn xt:Cc bn th ngh xem ti sao khng xt hm s

3

2 2x xf x ?

Trong mt s bi ton BT hai bin ta phi bin i c lp mi bin v

mt v, khi xut hin hm c trng cn kho st.

2. Dng 2: Kt hp vi cc BT khc nh BT AM-GM, BT

Cauchy-Schwarz, BT Chebyshes,

i vi cc bi ton phc tp, ta cn phi hp vi phng

php chn khong cc bin v cc BT ph khc nh BT AM-GM, BT

Cauchy-Schwarz, BT Chebyshes, hoc cc nh gi khc , hoc phi hp

vi cc phng php khc nh phng php ta , ....

Ta thng c lng T(x, y, z, ...) bi mt hm s ch ph

thuc vo mt bin s, t kho st hm s ny t c mc ch.


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40

V d 20: Cho cc s , , 0;1x y z tha mn 1 1 1xyz x y z .

Chng minh rng

2 2 2 3

4x y z .

Gii:

Ta c

1

2 1

xyz x y z xy yz zx xyz

xy yz zx xyz x y z

M

2 2 2 2

22 2 2

2

2 2 4

x y z x y z xy yz zx

x y z x y z x y z xyz

p dng BT AM-GM ta c3

3

x y zxyz

, suy ra

3

22 2 2 2 2 43

x y zx y z x y z x y z

.

t t = x + y + z (0 < t < 3), ta c3

2 2 2 2 42 227

tx y z t t .

Kho st hm s 3

2 42 227

tf t t t

ta c 3

min

4

f t khi1

2

x y z .

V d 21: ( Tuyn sinh i hc Vinh, 2001)

Chng minh rng nu a, b, c l di ba cnh ca mt tam gic c chu

vi bng 3 th

2 2 23 3 3 4 13 1a b c abc .

Gii: t2 2 23 3 3 4T a b c abc . Do vai tr ca a, b, c bnh ng

nn khng gim tng qut ta c th gi s 0 a b c .


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41

T a + b + c = 3 v a + b > c suy ra3

12

c (2).

Ta bin i

22 2 2 2

2 2

3( ) 3 4 3 2 3 4

3 3 3 2 3 2

T a b c abc a b ab c abc

c c ab c

Do 2 3c > 0 v 2

32

a bab

, suy ra

2 22

22 2

3 2

13 3 3 3 2

2

13 6 9 3 3 3 22

3 27

2 2

T c c a b c

c c c c c

c c f c

Ta c 23 3f c c c , nn f(c) ng bin trn 31;2

. V vy

1 13T f c f .

ng thi 13 1T c . Vi gi thit 0 a b c v a + b + c = 3v (3) suy ra a = b = 1, tc l tam gic ABC u.

3. Dng 3: Kho st hm s theo tng bin

i vi cc BT nhiu bin, ta c th chn mt bin l bin s

bin thin v c nh cc bin cn li, bi ton lc ny tr thnh BT mt

bin.

V d 22: Cho0 , 1a b . Chng minh rng

tan . tan tana b ab .

Gii: Gi s a b . t tan .tan tanf x b x bx vi 1b x . Ta c

2 2tan

os os

b bf x

c x c bx .

Do 0 , 1a b nn tan 0b b v 2 21 1

os osc x c bx suy ra 0f x ,

nn f ng bin trn [b; 1]. V vy vi a b ta c f a f b . Suy ra


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42

2 2tan .tan tan tan tan 1a b ab b b .

t 2 2tan tang x x x , c

22tan 2

0; 0;1cos os

x x

g x xx c x

Suy ra 2 20 0 tan tan 0 2g b g b b

T (1) v (2) suy ra pcm.

V d 23: Chng minh rng

3 3 3 2 2 22 3, , , 0;1x y z x y y z z x x y z .

Gii: BT cho tng ng vi

3 2 2 3 3 22 2 3f x x yx z x y z y z .

Ta c

2 26 2f x x yz z v

2 21

2 22

16

601

66

x x y y zf x

x x y y z

.

V 0x nn 1 0;1x . Xt hai trng hp

Nu 2 0;1 0, 0;1x f x x . Suy ra f(x) gim trn [0; 1].

Do

0;1ax ax 0 , 1

xm f x m f f

.

Nu 2 0;1x th ta c bng bin thin

x 0 2x 1

f - 0 +

f

2f x


49/68

43

T bng bin thin suy ra

0;1ax ax 0 , 1

xm f x m f f

.

Nh vy trong c hai trng hp ta u c

0;1ax ax 0 , 1

xm f x m f f

.

Mt khc

3 3 2 3 3 2 20 2 2 2 1f y z y z y z y z y z f .

Ta s chng minh 1 3f . Tht vy, t

3 3 2 21 2 2f g y y z y z y z

Ta c

21

2

22

16 0

66 2 1 01

66

y y z zg y y zy

y y z z

Nu 2 0;1 0, 0;1y g y y . Suy ra g(y) gim trn [0; 1]. Do

0;1ax ax 0 , 1

ym g y m g g

.

Nu 2 0;1y th ta c bng bin thin

y0 2y 1

g - 0 +

g

2g y

T bng bin thin suy ra

0;1

ax ax 0 , 1y

m g y m g g

.


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44

Nh vy trong c hai trng hp ta u c

0;1ax ax 0 , 1

ym g y m g g

.

Ta c

3 2 3 20 2 2 2 2 1 1 1 2 1 3 3g z z z z z g z z z

Vi mi 0;1z .

2.2.2. Rn luyn k nng s dng hm s n iu chng minh bt

ng thc.

chng minh bt ng thc, ngoi cc bt ng thc kinh innh bt ng thc Cauchy, bt ng thc Bunhiacopxki ..., th s dng o

hm cng l mt cng c hu ch. Trong nhiu trng hp, s dng phng

php kho st hm s chng minh bt ng thc th li gii bi ton s

ngn gn v n gin hn rt nhiu.

Gi s ta cn chng minh bt ng thc : A B trn tp D (vi D l

mt on, khong, na on hay na khong).

S dng o hm chng minh bt ng thc ta thng dng hai

cch sau:

Cch 1: Xt f l mt hm s ca mt i s no , f xc nh trn tp

D v tha mn f() =A, f()=B , vi , D v f n iu trn D.

Nu chng minh f (x) nghch bin trn D

Nu , chng minh f (x) ng bin trn D.Cch 2: Xt hiu f=A B trn D v coi y l hm s ca mt i s

no .

Nu f nghch bin trn D,

cn ch ra tn ti: , D : , f() A B v f() = 0

A B.


51/68

45

Nu f ng bin trn D, cn ch ra tn ti : , D, , f()A

B v f()=0 A B.

Cch 2 thc cht l mt trng hp ring ca cch 1.

Xt cc v d sau:

V d 24.Cho 0 < x 0

f(x) nghch bin trn khong (0 ; +)

Do f(x)< f(0) vi x>0

x2

6

x sinx 0

x 2

6

x0 (pcm)

Nhn xt: cc v d trn ta u nhn ra ngay c hm s cn xt s

bin thin. V d tip theo minh ha vic la chn hm s thch hp xt s

bin thin, khi bt ng thc cn chng minh l bt ng thc nhiu bin.

V d 26.Chng minh rng nu x + y = 1 th 4x +4y 1

8

Bi gii

T x+y = 1 suy ra : y = 1 x nn 4x + 4y = 4x + 41 x

Xt hm s f(x) = 4x + 4

1 x

Ta c f'(x)= 4. 33 4 1x x

f'(x)= 0 x= 12

Bng bin thin

x - 12

+

f(x) - 0 +

f(x)

1

8


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47

Da vo bng bin thin ta c f(x) >18

vi mi x

ng thc xy ra khi x=y= 12

V d 27.Chng minh rng trong mi tam gic ABC nhn ta u c:

2

3(sinA+ sinB+ sinC ) +1

3(tanA+ tanB+ tanC ) >

Bi gii

Do A + B + C = nn bt ng thc cn chng minh tng ng vi

bt ng thc:

(23

sinA +13

tanA A ) + (23

sinB +13

tanB B ) +

+ ( 23

sinC +13

tanC C) 0

Xt hm s f(x) 23

sinx +13

tanx x vi 0


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23

sinC +13

tanC C > 0

(23

sinA + 13

tanA A ) + (23

sinB +13

tanB B ) + (23

sinC +13

tanC C ) > 0

(sinA+ sinB+ sinC ) +13

(tanA+ tanB+ tanC ) > (pcm)

Nhn xt: Trong cc v d trn ta da vo cc iu kin t gi thit

bin i ri t la chn hm s thch hp xt tnh n iu.Tuy nhin,

trong nhiu trng hp ta phi s dng thm cc bt ng thc quen thuc

nh bt ng thc Csi, Bunhiacopxki2.2.3. Rn luyn k nng chng minh bt ng thc v tm GTLN-GTNN

ca biu thc bng phng php hm s

V d 28:Cho x, y, z l cc s thc khng m tho mn 1x y z .

Tm gi tr ln nht ca biu thc:9

4P xy yz xz xyz

(T kt qu ca bi ton 2.2.3.1, ta d dng tm c gi tr ln nht

ca biu thc P l1

4).

+) Tip theo yu cu HS tm gi tr nh nht ca biu thc P

(Da vo bt ng thc ( )( ) 9x y z xy yz zx xyz ta d dng th

tm c gi tr nh nht ca biu thc P bng 0).

Nh vy ta gii c bi ton:

V d 29 .Cho x, y, z l cc s thc khng m tho mn 1x y z .

Tm gi tr ln nht, gi tr nh nht ca biu thc A mxyz xy yz zx ,

trong m l s thc cho trc.

(p s:

Nu 9m th gi tr ln nht ca A l1

4,


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49

gi tr nh nht ca A l9

27

m .

Nu9

9

4

m th gi tr ln nht ca A l1

4

, gi tr nh nht ca A

l 0.

Nu9

4m th gi tr ln nht ca A l

9

27

m , gi tr nh nht ca A l

0).

T ta c cc bt ng thc:

Nu x, y, z l cc s thc khng m tha mn 1x y z th:

a)9

27

m mxyz +xy +yz + zx

1

4

( Trong m l s thc cho trc v 9m ).

b) 0 mxyz +xy +yz + zx 1

4

(Trong m l s thc cho trc v9

9

4

m ).

c) 0 mxyz +xy +yz + zx 9

27

m

(Trong m l s thc cho trc v9

4m ).

T cc bt ng thc trn ta c th xy dng c cc bi ton mi sau:

V d 30. Cho x, y, z l cc s thc khng m tho mn 1x y z .

Chng minh rng7

0 227

xy yz zx xyz

V d 31. Cho x, y, z l cc s thc khng m tho mn 1x y z .

Chng minh rng 3 3 32

3 19

a b c abc

c bit ha bi ton 3, bng cch thay gi thit bi ton tng qut bi gi

thit: x, y, z l di ba cnh ca tam gic c chu vi bng 1, ta c bi ton:


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50

V d 32. Cho x, y, z l di ba cnh ca tam gic c chu vi bng 1.

Tm gi tr ln nht, gi tr nh nht ca biu thc: 2P xy yz zx xyz .

(P khng c gi tr nh nht)

T v d 31d dng suy sau:V d 33.Nu a, b, c l di ba cnh ca mt tam gic c chu vi bng

1 th1

2 274

ab bc ca abc ra v d

T (4) v v d 6 ta c v d sau:

V d 34. Cho tam gic ABC c chu vi bng 1. Chng minh rng

3 3 32 139 4a b c abc .

2.3. Rn luyn mt s k nng gii bt phng trnh theo quan im hm.

2.3.1.Rn luyn k nng gii bt phng trnh bng phng php vn

dng khi nim TX v tp gi tr.

Khi gii mt BPT ta c gng bin i n v mt BPT tng ng m

cc biu thc m hay logarit cng c s, sau ly m ho hoc logarit ho

cc v kh biu thc m hoc logarit cha n s, i vi BPT logarit cnch t iu kin biu thc c ngha.

Ta xt mt vi v d thy c tm quan trng ca TX

V d 35: Gii bt phng trnh sau: 2x xlog (3x - 1) > log (x + 1) (1)

Bi gii:

2

2

2

2

x > 1 x > 1x > 1

1< x x + 1 1< x 00 < 3x - 1 < x + 1

2 1x 3 2 0

x

xx

x

x

x xx


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51

Vy nghim ca BPT l 1

x ( ; 2) \ 13

+ Cng c th c cch trnh by khc nh sau:

2

2

2

0 10 1

3x - 1 > 0 1(1) 1/ 3

x + 1 > 0 1/ 3 2(x - 1)( x + 3x + 2) > 0

(x - 1)(3x - 1 - x - 1) > 0

x

x

x

x

x

Vy nghim ca BPT l 1

x ( ; 2) \ 13

.

Ch :Gii bi ton BPT bng PP bin i tng ng i hi hc

sinh phi c s cn thn v t m cao, bi ton lun yu cu s chnh xc, y

cc iu kin sao cho cc bc bin i l tng ng. c th rn luyn

cho HS kh nng gii nhiu bi ton khc, l bi ton tm TX ca hm

s, bi ton v s bin thin ca hm s, bi ton gii BPT...

V d 36: Gii BPT: 2log (5 8 3) 2x

x x (1).

Bi gii:

(1)

2

2 2

2

2 2

2

11 34 8 3 0

5 8 3 20 11 30 1

5 8 3 0 2 50 5 8 3

4 8 3 0

xx

x x xx x xx

xxx x

x x x

x x

Vy BPT c nghim: 1 3 3( ; ) ( ; )2 5 2

x .

Ch : bi tp 2, nhiu HS mc phi sai lm c bn, HS thng

khng ch n iu kin ca c s a nn vn dng sai tnh cht n iu

ca hm s logarit, nhiu em gii bi tp nh sau:


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52

iu kin: 23

5 8 3 0 51

xx x

x

Khi :

2log (5 8 3) 2x x x 2 2 21

25 8 3 4 8 3 03

2

xx x x x x

x

Vy BPT c nghim: 1 3( ; ) ( ; )2 2

.

GV cn lu cho HS sai lm c bn ny, lu vi cc em rng:

Bn cht:( )

( )

( )

1

log0 1

0

bx

a x

bx

a

f af b

a

f a

2.3.2. Rn k nng gii bt phng trnh thng qua s dng tnh n

iu ca hm s

- nh ngha : Cho hm s y = f( x) xc nh trong khong (a, b ).

+ f tng ( hay ng bin ) trn khong (a, b ) x1, x2 (a, b) : x1< x2

f( x1) < f (x2) .

+ f gim ( hay nghch bin ) trn khong (a, b )

x1, x2 (a, b) : x1< x2

f( x1) > f (x2) .

- Cc tnh cht :


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54

(2) f (t) > f (1) t < 1

log4x < 1. 0 < x < 4.

V d 38. Gii cc bt phng trnh sau :

a, 5x + 2 3x < 9

b, 3 3 2x +5

2 1x - 2x 6.

c, 2x 2x 3 - 2x 6x 11 > 3 x - 1x

+ Nhn nh :Cu a, bn hon ton c th s dng phng php bnhphng hoc bin i tng ng gii. Tuy nhin, ti mun hng bn

n vic s dng tnh n iu ca hm s gii quyt, tuy nhin on c

mt nghim ca phng trnh ny mt kh nhiu thi gian (bn ch chn

nhng s sao cho biu thc di du cn l s chnh phng).

Cu b, c th t n ph, nhng bin i kh di. Bi ton n gin nu

s dng tnh n iu ca hm s gii bt phng trnh trn s ti u hn .

Cu c, kh phc tp v cng c th t n ph. Song nu quan st k th

thy c mi quan h tng t gia cc thnh phn trong bt phng trnh.

Li gii :

a, iu kin : x - 32

.

Bt phng trnh cho tng ng vi 5x + 2 3x - 9 < 0.

Xt hm s f (x) = 5x + 2 3x - 9 trn [ -32

, + ).

Ta c hm s f(x) l hm s lin tc v c o hm


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55

f (x) =1

2 5x +

1

2 2 3x > 0 x (- 3

2 , + )

nn f( x) ng bin trn (- 3

2

, +).

Hn na f (11) = 16 + 25- 9 = 0, v vy min nghim ca bt

phng trnh cho phi tha mn3

2( ) (11)

x

f x f

3

211

x

x

Kt lun : Tp nghim ca bt phng trnh l [ - 3

2

, 11 )

b, iu kin 12

< x 32

t f (x) = 3 3 2x +5

2 1x - 2x .

Khi f (x) lin tc trn ( 1

2

, 3

2

] v

c o hm f (x) = -3

3 2x +

5

2 1x - 2 < 0 x ( 1

2 , 3

2)

nn n ng bin trn ( 12

, 32

)

Hn na f (1) = 6.

Do o tp nghim ca phng trnh phi tha mn :

1 3

2 21

x

x

1 x 32

Kt lun : tp nghim ca bt phng trnh l [ 1, 3

2

].


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56

c, iu kin : 1 x 3.

Bt phng trnh cho tng ng

2x 2x 3 + 1x > 2x 6x 11 + 3 x

2x 1 2 + 1x > 23 x 2 + 3 x (1)

t u = (x 1)2, v = (3 x)2, th (1) tr thnh 2u + u> 2v +

v

Hay f (u) > f (v).

Xt hm s f(t) = 2t + ttrn [1, 3]

Khi f( t)l hm lin tc trn [ 1, 3] v c o hm

f( t) =1

2 2t +

1

2 t> 0 trn [1, 3] nn f (t) ng bin trn [1;3].

V tnh ng bin nn t f(u) > f( v) suy ra u > v hay x 1 > 3- x x> 2

Kt hp vi iu kin ta c 2 < x 3.

Kt lun tp nghim ca bt phng trnh l ( 2, 3].

2.3.3. Rn k nng gii bt phng trnh bng phng php s dng o

hm v xt s bin thinVn p dng o hm, tnh n iu ca hm s gii BPT ngi

ta dng t lu, nhng trong SGK t c bi ton p dng cch gii ny, cho nn

hc sinh khng quen dng v k nng dng cha c tt.

Cc nh l c bn


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57

nh l 1: Nu hm s y = f(x) lun ng bin (hoc lun nghch bin)

v lin tc trn D th s nghim ca phng trnh f(x) = k (k l s khng i)

trn D khng nhiu hn mt nghim v f(x) = f(y) khi v ch khi x = y vi

mi x, y thuc D. Chng minh:

Gi s phng trnh f(x) = k c nghim x = a, tc l f(a) = k. Do f(x)

ng bin nn

x > a suy ra f(x) > f(a) = k nn phng trnh f(x) = k v nghim

x < a suy ra f(x) < f(a) = k nn phng trnh f(x) = k v nghim Vy pt

f(x) = k c nhiu nht l mt nghim.

Ch : T nh l trn, ta c th p dng vo gii phng trnh nh sau:Bi ton yu cu gii pt: F(x) = 0.

Ta thc hin cc php bin i tng ng a phng trnh v dng

f(x)=k hoc f(u) = f(v) ( trong u = u(x), v = v(x)) v ta chng minh c

f(x) l hm lun ng bin (nghch bin) Nu l pt: f(x) = k th ta tm mt

nghim, ri chng minh l nghim duy nht.

Nu l pt: f(u) = f(v) ta c ngay u = v gii phng trnh ny ta tm cnghim. Ta cng c th p dng nh l trn cho bi ton chng minh phng

trnh c duy nht nghim.

nh l 2: Nu hm s y = f(x) lun ng bin (hoc lun nghch bin)

v hm s y = g(x) lun nghch bin (hoc lun ng bin ) v lin tc trn D

th s nghim trn D ca phng trnh: f(x) = g(x) khng nhiu hn mt.

Chng minh:Gi s x = a l mt nghim ca phng trnh: f(x) = g(x), tc l f(a) = g(a).

Ta gi s f(x) ng bin cn g(x) nghch bin.

Nu x > a suy ra f(x) > f(a) = g(a) > g(x) dn n phng trnh f(x) =

g(x) v nghim khi x > a.

Nu x < a suy ra f(x) < f(a) = g(a) < g(x) dn n phng trnh f(x) =

g(x) v nghim khi x < a.


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58

Vy pt f(x) = g(x) c nhiu nht mt nghim.

Ch : Khi gp phng trnh F(x)=0 v ta c th bin i v dng

f(x)=g(x), trong f(x) v g(x) khc tnh n iu. Khi ta tm mt nghim

ca phng trnh v chng minh l nghim duy nht.Nhn xt:

+ Nu hm s f(t) ng bin trn tp D, th f(x) >f(y) x > y (vi x, y

thuc D)

+ Nu hm s f(t) nghch bin trn tp D, th f(x) > f(y) x < y (vi

x, y thuc D)

Ta s xt mt vi v d thy c vai tr ca xt s bin thin

V d 39:

Tm tham s m bt phng trnh sau c nghim:

3 1mx x m (1)

Gii

iukin: 3x . t 23 , 0 3t x t x t

BPT (1) trthnh 2

2

1

( 3) 1 2

t

m t t m m t

(2)

Xt hms2

1( )

2

tf t

t

, 0t

Ta c:

2' ' 2

22

1 32 2( ) , ( ) 0 2 2 0

1 32

tt tf t f t t t

tt

Bng bin thin

x - 3 1 +

f(x) - 0 +

f(x) 1 34

1

2

0


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59

Suy ra

0;+

1 3ax

4m f t

Bt phng trnh (1) c nghim 3x bt phng trnh (2) c

nghim 0t

0; )m ax ( )m f t

3 1

4m

* Nhn xt:

Nu a v bt phng trnh 1 33 11

xmx x m m

x

.

Khi hm s: 1 31

xf x

x

dn n vic tnh o hm, gii phng

trnh o hm v xt du o hm gp kh khn.

2.4. Kt lun chng 2.

Chng ny trnh by vic rn luyn k nng gii bi ton BT v

BPT cho hc sinhTHPT theo quan im hm s, bao gm:

Rn luyn k nng gii bt phng trnh bng phng php vn dngkhi nim TX v tp gi tr.

Rn k nng gii bt phng trnh thng qua s dng tnh n iu

ca hm s.

Rn k nng gii bt phng trnh bng phng php s dng o

hm v xt s bin thin.


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60

KT LUN

i chiu vi mc tiu, nhim v v kt qu nghin cu trong qu

trnh thc hin ti:Rn luyn k nng gii bt phng trnh v btng thc cho HS trung hc ph thng theo quan im hm s chng

em thu c kt qu sau:

Trn c s phn tch cc cng trnh ca nhiu nh khoa hc trong v

ngoi nc, ti h thng ha cc quan im v k nng, quan im hm

s t a ra mt s bin php rn luyn k nng gii bi ton bt ng

thc v bt phng trnh cho hc sinh THPT.


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61

TI LIU THAM KHO

1. G.PoliA(1997), Gii bi ton nh th no,NXB Gio dc, H Ni

2.

Nguyn Trng Bo, Nguyn Huy T (1992), Ti nng v chnh sch i

vi nng khiu, ti nng, Vin khoa hc Gio dc, H Ni.

3. Trn nh Chu (1996), Xy dng h thng bi tp s hc nhm bi

dng mt s yu t nng lc ton hc cho hc sinh kh gii u cp

THCS, Lun n Ph tin s khoa hc S phm Tm l, Vin khoa hc

Gio dc, H Ni.

4. Hong Chng (1997), Nhng vn logic trong mn ton trng ph

thng THCS, Nxb Gio dc, H Ni

5.

Nguyn c ng, Nguyn Vn Vnh(2001), Logic Ton, Nxb Thanh

Ha, Thanh Ha.

6. T in ting Vit (1997), Nxb Nng v Trung tm T in hc, H

Ni - Nng.

7.

Nguyn Th Hng Nga 2011, Rn luyn k nng gii ton tm gii hn

trong chng trnh lp 11 THPT (Ban c bn)

8.

Cruchetxki V. A. (1973), Tm l nng lc ton hc ca hc sinh,Nxb Gio

dc, H Ni.

9.

Nguyn B Kim(2002),Phng php dy hc mn ton,Nxb S phm H

Ni, H Ni

10.

o Vn Trung(2001),Lm th no hc tt ton ph thng, Nxb i

hc quc gia, H Ni.

11. o Vn Trung (1999), Nhng vn c bn gio dc hin i, Nxb

Gio dc H Ni.

12.

L Don T, T Duy Hp (2002), gio trnh Logic hc, Nxb chnh tr

Quc gia, H Ni.

13.Nguyn Cnh Ton (1997), Phng php lun duy vt bin chng vi

vic hc, dy, nghin cu ton hc, tp 1, Nxb i hc Quc gia H Ni.


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14. Phm Vn Hon, Nguyn Gia Cc, Trn Thc Trnh (1981), Gio dc

hc mn Ton, NXB Gio dc, H Ni.

15.

Nguyn Hu Lng (2002), Dy v hc hp qui lut hot ng tr c,

NXB Vn ha thng tin, H Ni.

16.

G. Polya (1997), Ton hc v nhng suy lun c l, NXB Gio dc, H

Ni.

17. Phan c Chnh, Phm Vn iu, Vn H, Phan Vn Hp, Phm

Vn Hng, Phm ng Long, Nguyn Vn Mu, Thanh Sn, L nh

Thnh (2000), Mts phng php chn lc gii cc bi ton s cp tp

2, NXB i hc quc gia H Ni, H Ni.18.Sch gio khoa Ton 10 Chng trnh c bn, NXB Gio dc, H Ni.

19.

Tp chTon hc v tui tr,NXB Gio dc, H Ni.

20.

Tuyn chn theo chuyn Ton hc v tui tr, Quyn 6, NXB Gio

dc, H Ni.

21.Mt s website tham kho

Http://www.dienantoanhoc.net/Http://www.vnmath.com/

Tp chTon tui th,NXB Gio dc, H Ni.

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