1. Trang chủ
  2. » Tất cả

Chuyển từ ngôn ngữ đại số sang ngôn ngữ số học trong việc hướng dẫn học sinh tiểu học giải các bài toán có lời văn

3 2 0

Đang tải... (xem toàn văn)

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 3
Dung lượng 163,48 KB

Các công cụ chuyển đổi và chỉnh sửa cho tài liệu này

Nội dung

CHUYEN TIT NGON NGIT DAI sH SANG N G O N N G V S H HOC IRDNG VKC HUONG DJlN HOG SINH TilU HDG GlAl CHG BAl TDUN Gll III! VAN O ThS T H A I H U Y VINH" Bdng / I ng dfin cho hqc sinh tllu hqc (HSTH) kha[.]

Trang 1

CHUYEN TIT NGON NGIT DAI sH SANG N G O N N G V S H HOC

IRDNG VKC HUONG DJlN HOG SINH TilU HDG GlAl CHG BAl TDUN Gll III! VAN

O ThS T H A I H U Y V I N H "

Bdng /

Iihdc cdc cdch gldl bdl todn cd Idl vdn Id ng dfin cho hqc sinh tllu hqc (HSTH) khai

mdt bl^n phdp hChj hl^u trong v l ^ rfin luy$n

vd phdt trlln ngdn ngiJ (NN) todn hgc cho hqc

sinh {HS) Gidl cdc bdl todn cd Idl vdn thyc ch^t

Id chuyin h> N N thdng thudng song N N vd ki

hi|u todn hqc Trong khudn khd bdl vilt ndy, chung

tdl dua ra mdt sd bdi todn cy t h l vd Irlnh bdy

phuong phdp khai thdc cdc cdch gldl

khde nhau bdng cdch ldp phuong

h'inh bde nhdt 1 dn vd h§ phuong h'inh

bde nhdt 2 d'n Sou dd, chuyin djch

h> N N dqi sd' song N N sd' hqc d l

hudng dfin cho HSTH gldl theo N N

sd hqc nhdm cung cd vd phdt triin

ndng lye \u duy, N N todn hqc eho

cde em

Bdi todn 1 (bdi todn cd d Kiu

hgc (TH)): "Vua gd vua ehd, bd Igi

eho trdn, ba muoi sdu eon, mgt trdm

chdn chdn Hdl cd bao nhliu con gd,

bao nhiiu con chd?"

O d d y , chung tdl mudn khai thdc

td't ed cdc cdch gidi dql sd bdng edeh

ldp phuong H n h bde nhd't 1 d'n sdvd

hfi phuong h'inh bde nhd't 2 dn sd';

sou dd, ehuyen djch sang N N sd' hqc

d l huong dfin cho HSTH

1) Gidi vd khai thdc bdl todn 1

bSng cdch lip phuong trinh bic

nhd't 1 d'n vd hi phuong trinh bic

nhdt 2 dn sd' Trudc hit, ta cdn xdc

djnh bdl todn cd mdy dgl lugng chua

bilt? (dd Id sd' con gd, sd' chdn gd,

sd'con chd, so'chdn end) Chf cdn bilt

1 Kong 4 dgl lugng sfi Km duge cdc

dqi luqng cdn Iqi

To ed cdc cdch gldl bdng N N dql

sd duqc Ktnh bdy trong bdng dudi

ddy (xem bdng 1)

hfong i}ng vdl 4 c 6 ^ gldl bdng cdch ldp phuong h'inh bde nhd^ 1 dn sd, 6 cdch ehqn 2 dn » hnhg i>ng vdl d cdch gldl bdng cdch Idp h | f r f i u ^ i ^ h-Inh bde nhdt 2 6n s^ (d ddy quan nl|m mSl ci^cn^ chqndnsdid 1 cdch gidi, Kong mfil cdch ehqn an

sd sfi cd nhllu cdch bllu dlln phuong Kinh hoy h$

phuong Ktnh khdc nhau md d bdng dudi ddy

chung Idl ehl mdl nfiu 1 cdch bllu d l l n

T T

1

2

3

'

S

s

7

e

9

10

con at

«

2

3e-x

3 6 - i

2

2

3e-x

si 2i<

2(36->i) 100-X

y

2

2x

100-y

S&

a e - x tOO-j

X

1

X-tL

y

>

'

4

Si diAncM

MX-n)

l W - ) i

4x

100-y

* y

y

100-x

y

y

"*^isiiirb2"*

2x « 4(30 •>)-100

x - 2 2

2 J 2(36-«)*4i = 100

x * 2 ( 3 6 - - ) -100

x = 5e

=9X-22:v-44

(2x**y-tM

lx+y-36

X = 22 V > 14 l2r*>-IOO

| i - - 1 0 0

I ' 2 2 y - 5 6

If*""

x = 44;y = 56

x = 14:v«96

Nhi/vqy, t>d/laSn ) c 6 4 c a c h c h 9 n 1 Aistf ' S i C t t t l i e i i M l l H l W I i

Tap chi Blip due s6 2 9 8 (M a n/»oi»i

Trang 2

gidl tnqt bdl todn bde nhd't bdng cdch ldp phuong

trinh bqc nhd't 1 d'n sd vd hg phuong trtnh bde

nhdt 2 dn sd nhu sou: ne'u cd n dql lugng ehua

bilt thi sfi cd n edeh gldi bdng edeh ehgn 1 dn sd

vd ed C^„ cdch gidl bdng edeh ehqn 2 dn sd, do

dd, to ed n -t- C^^ cdch

2) Chuyin td cdch gidl bdng NN dgt sd

sang NN sd hgc di hudng ddn eho HSTH:

Cdch 1 • G i d su 3 d con d i u Id chd ed, N h u

v^y, mdi eon gd da «thgm vdo" 2 chdn Luc d d ,

t^ngsdchdn Id: 3 d x 4 = 144 (chdn) S d e h d n g d

dd «thgm vdo" Id: 144 - 1 0 0 = 4 4 (chdn) Suy ra

sd,gd Id: 44; 2 = 2 2 (eon); sd chd Id: 3 d - 22 =

14 (con)

Cdch 2: G i d si> 3 d con deu Id gd ca Khi d d ,

moi eon ehd dd «bdt d i " 2 chdn Luc d d , td'ng sd'

chdn Id: 3 d x 2 = 7 2 (chdn) So chdn chd dd «bdt

di" Id 100 - 7 2 = 28 (chdn) Suy ra sd chd Id:

28 : 2 = 14 (con); so g d Id: 3d - 14 = 22 (con)

Cdch 3: G i d su 100 chdn deu Id chdn eh(^

khi dd mol con gd dd "thfim vdo" 2 chdn Vdy,

tong so' con vdt Id: 100: 4 = 25 (eon) Sd' eon vdt

«hyt" di Id: 3 d - 25 = 11 (eon) Suy ra so gd Id:

11 x 2 = 2 2 (eon); sd chd Id: 3d - 22 = 14 (eon)

Gido vidn (GV) ed t h l ed cdch li gldl khdc

cho HS nhu: Id'y 3d cdi hjl tuqng trung cho 3d

con vdt vd 100 hqt ddu tuqng trung cho 100 cdi

chdn Gid su 100 chdn deu Id chdn chd thi phdl

bd mol tui 4 hqt vd bd duqc 25 hii Cdn 11 tul

chua ed hqt ndo, budc phdi bdt 22 hqt d 11 Kil

dd bd 4 hqt ddu, mol hji 2 hqt Luc d d , ha se cd

n X 2 = 22 KJI d y n g 2 hqt ddu vd 14 tul dyng 4

hgt ddu N h u vdy, gd cd 22 con, chd cd 14 con

Cdch 4: G i d su 100 chdn deu Id so chdn gd

thi khi do, moi eon chd d d "bdt d l " 2 chdn Vdy,

tong sdcon Id: 100 : 2 = 5 0 (con) Sdeon "tdng

thgm" Id: 50 - 3d = 14 (con) Suy ra so' chd Id 14

con vd so' gd Id: 3 d - 14 = 22 (con)

GV If gidi cho HS tuong tu nhu cdch 3: Idy 3d

cdi hji hrong h v n g cho 3 d con vdt vd 100 hqt

ddu hjgng h'ung cho 100 cdi chdn Gid su tdt cd

100 chdn deu Id chdn g d , khi bd mSl tul 2 hqt thi

cdn phdi cd 5 0 tul Luc d d , t h i l u 14 tul vd cd 28

hgt ddu chua duqc bd vdo tul ndo Bd mol tui

thfim 2 hqt nua se cd 14 hjl dyng 4 hqt ddu Suy

ra sd chd Id 14 con vd sd gd Id 2 2 con

Cdch 5: GV hudng dan HS ve so dd doqn

tndng nhusau:

Gid su sd' chd it hon sd g d , to cd so dd:

Tap chi Glao duc so 2 9 8 (fcia-n/aoiai

S d e h d n g d (II): I 1 1 Sdcon ehd llll): ,

Sd'ehdnehd(IV); ^ [ '

T a c d : ( l ) + {lll) = 3d;[ll) + ( I V ) = 1 0 0 Quo so d d , HS d l ddng thdy ngay 2 Idn sd chd Id: 100 - (3d X 2) = 28 (con); sd chd Id:

28 : 2 = 14 (eon); s6 gd 3d - 14 = 22 (con)

GV ed the chuyin so dd doqn Kidng thdnh so

dd khde khdl qudt hon md khdng cdn gid su sd ehd ft hon hay sd gd it hon sd ehd:

Bllu thj O Id sd gd thl sd chdn gd Id O O Bllu thj D Id sd ehd thi sd chdn chd Id D D D D _ , / idv

To ed so do: f,.^

D n

Dyo vdo so 66, HS de ddng thd'y ngay hai

hinh vudng ung vdi: 100 • 3d x 2 = 28 (con) Mdt hinh vudng ung vdi: 2 8 : 2 = 1 4 (con)

So chd 14 con, so gd Id 3d - 14 = 22 (con)

Ti> cdch I vd cdch 2, GV hudng dan HS gid

su cd so chd hodc so' gd Id mdt so n bd't ki, tuy nhidn, n phdi nhd hon hodc bdng 3d Chdng hqn, gid su so chd cd 3 con, khi dd sd gd Id: 3d - 3 =

33 (con} So chdn ehd Id: 3 x 4 = 12 So chdn gd Id: 33 X 2 = d d Td'ng so chdn gd vd chdn chd Id:

12 + d d = 78 Sdchdn "hyt" d i : 100 - 78 = 22 So' con gd p h d i thay b d n g sd con ehd I d :

22 : [4 • 2} = 1 1 Vdy, sdeon ehd Id: 3 + 11 = 14 (con); so con gd Id: 33 - 11 = 22 (con)

Gid su sdchd Id n eon (n < 3d), to ldp bdng sou:

0

D

i

0

D

i

st chd

n = 36

S i conga ( 3 6 - n )

36

34

2

0

S i

CM

n

1?

i f l f i

i^n

144

5 6

gi

IV

7n Hfi

0

Chan ChA

72

76

140 ItX)

S i chfln Iflng mam 2B

26

22

40

44

S i con gfl d i r ^ c

Ihay bin s i con

c h i Hoflc s i con

c n i Oirc^ tnev

Wn a i con gi

12

20

22 S6

14

S i

22

V7

22

22

22

HS ed the gid su so chd Id n, vdi 0 ^ n ^ 3d (n = 0 chinh Id cdch gid su 3 d con deu Id g d ,

n = 3 d Id cdch gldl gid su 3 d eon d i u Id chd cd), hodc gid siJ sd con gd h> 0 d i n 3 d

#

Trang 3

• trinh bde nhdt 1 d'n ho^c h§ phuong Irlnh bde

nhdt 2 d n , ta ed t h l Hm ro n h l l u cdch gldl sd hqe

khde nhau

Bd I t o d n 2 : Mit xe md td dg djnh dl h> tinh A

din tfnh B trong mit ^dl glan nhdt djnh Niu xe

chgy vdl vin tdc 35 km/ gid thi din nol chim 2

gid Niu xe chgy vdl vin tdc 50 km/gid thi din

nai sdm hon ] gid Tfnh qudng dudng AB vd thdi

gian dt/ dinh lOc ddu [Di thl ehqn HS gldl TH Knh

Ngh§ A n ndm hqe 2 0 0 0 - 2001)

Tuong ty nhu bdi todn 1, bdl todn ndy ed 5

dql luqng chua b i l l , dd Id: qudng dudng AB,

Kidl gian dl vdl vdn tdc 35 k m / g l d , thdi glan dl

vdl vdn tde 50 k m / g l d , thdi glan d y djnh luc ddu

vd vdn tdc d y djnh lue ddu N h u vdy, s3 cd 5

cdch chon 1 d'n sd, ldp phuong trinh bde nhdt 1

d'n sd vd C ' = 10 edeh ehqn 2 dn sd, ldp h$

phuong trinh bde nhd't 2 d'n sd, suy ra cd 15 edeh

gldl bdng N N dql sd D i n ddy, GV cd t h l hudng

dfin HSTH gldl bdl todn bdng N N sd hqe

Q u o hal bdl todn ey t h l d trgn cho thdy, ddl

vdl bdl todn ed Idi vdn duqc g l d l b d n g cdch

ldp phuong Kinh bqc nhd't 1 d'n hode h | phuemg

trinh bqc nhd't 2 dn sd, n l u cd n dql lugng chua

b i l t Kii se cd n cdch ehqn 1 dn sd rdl Iqp phuong

h'inh bde nhd't 1 dn vd cd C^'cdeh chon 2 dn sd

sou dd ldp hg phuong trinh trinh bde nhd't 2 d'n

N h u vqy, cd tdt cd n + C„^ cdch g l d l dql id Tt>

cdch gldl dql sd to c h u ^ n song cdch g l d l s ^

hgc d l hudng dfin eho HSTH; i>ng vdi mfil edeh gldl thl N N todn hqc duge trtnh b d y vd d i l n dqt cung khdc nhou Ddy Id mdt trong nhiJng bl§n phdp g d p phdn nfing coo chd't lugng dgy

hqc todn dTH.Q

Tdi li^u Iham khdo

i Ph9m Dinh Thyc M^t c&u hdi v& d&p vi vifc d^y

toiln * tiiu hpc NXB Cido dvc, H 2004

2 G Polya Gidi bdi todn nhir th£ ndo NXB Cido

due H 1997

3 G Polya, Sdng t^o todn hpc NXB Cido diic, H

1997

4 D 6 Trung Hi$u - VO Duong Thuy Nhiing phuvng

phdp gidi todn & ti^u hpc NXB Dpi hpc su phpm, H,

1980

5 Tr&n Dien Hiin 10 chuyfin d£ b i i diKhig hpc sinh

gidl Todn 4 • 5 NXB Gido due H 2003

SUMMARY

The solving problems in elementary school have the text Is very rich, diverse and unique So teachers have to use common knowledge ond language of lem then find a guide forstudents This article I raised the extraction method of solving the most - /rove

me text problem by addressing the most equation

of a system of equations and unknovms most two unknowns, from which teachers seeking to move from language to language algebra arithmetic to guide elementary students explore the solutions by numerical methods In accordance with the pro-gram

ITng dung E-leamIng trong

(Tlip thea trang 53}

- E)dnh gid vd p/idn hii ngudi hgc: Website

e-leorning phdl thudng xuy6n k l l m Ko qud trinh

K i p Kiu k i l n thue vd RLNVSP cOo SV, Vi§c Kinh

bdy bdl gidng cdn duqc d i l u chlnh theo ndng

lye vd l i l n d q hqc tqp eua SV Vua k i l m Ira thudng

xuy&n, vuo x u li kjp tbdl nhijng phdn hdl cua SV

v l lien do vd nhung vd'n d l phdt sinh

Vide ung d y n g e-learning trong RLNVSP cho

SV ngdnh su phqm todn se d u o ro mqt mdi

trudng ddo tqo mdl thfch hqp, uu vigt hem mdi

trudng d d o tqo t r u y i n thd'ng, d d Id Iqo mdl

trudng hogl ddng tieh eye, chu d d n g , sdng tqo

eho SV; ddm bdo ho trg SV r6n' luy§n KNDH

theo hudng phdn h o d Q

Tdl lifu tham khdo

1 Dko Tam (chO bifin) - Lfi Hiin Ducmg Tifp cfin cdc

phuong phdp d^y hpc khOng truyin thtfng trong day

bpc mtm Todn & trudng d^i hpc vd truong plil thdng NXB Dpi hpc suphpm H 2008

2 Trin Tnmg (chu biftn) - D^ng Xufin Cuong - NguySn

v a n H6ng - NguySn Danh Nam O'ng di^ng cAng ngh^

thdng tin vdo d^y tipc mOn Todn fr truimg phi thAng

NXB Gido due Viit Nam H 201 i

SUMMARY

Appik:a^on e-ieaming in practice ofpedago^: cai qualification forstudents of education mathemat-: Ics at the Universtles

This paper presents the role of pedagogic training activities forstudents of education mathematics at the university Also, It ^lowspos^lrty of e-ieoming and pro-poses a number of requirements tor the applk:atk}n of e-leamir)g in pedago0cc^ training forstudents of edu-catkxi mathematics In the directton ofrMerentkjtion

Tap chi filao due s6 2 9 8 (id a - n/aoia)

Ngày đăng: 18/11/2022, 12:27

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN

🧩 Sản phẩm bạn có thể quan tâm

w