Vận dụng phép biện chứng duy vật trong dạy học toán ở phổ thông

V A N DUN G PHE P BIE N CHUNG,DU Y VA T

TRON G DA Y HO C TOA N 6 PH O THON G

O TS. NGUYEN THANH HUNG' - NGUYEN TH| THU THUY"

Triet hoc duy vdt bien chung (THDVBC) dua

ra cac quy ludt chung nhdt ve sy phat

trien ty nhien, xa hoi vd tu duy cua con

ngudi; Id ca sd phuang phdp ludn giup con

ngudi hinh thdnh the gidi quan khoa hoc, Id

he thdng nhung nguyen tdc ve cdc quy ludt

khdeh quan nham dieu chinh hoat ddng nhdn

thuc cua con ngudi. Viec van dung cdc cap

phgm tru, cdc quy ludt cua THDVBC vdo dgy

hgc (DH) mdn Todn d phd thdng nhdm giup

hgc sinh (HS) phdt huy tinh tich cue, chu ddng,

sang tgo, ren luyen khd ndng ty gidi quyet vdn

de vd van dung todn hgc vdo thyc tien.

1. Mdt sd cap phgm tru ca bdn cua

THDVBC trong DH todn

1) Cdi rieng vd cdi chung: Id cap phgm

tru cd mdi quan he bien chung vdi nhau. Cdi

rieng Id cdi todn the, cdi chung Id cdi bd phdn

nen cdi rieng phong phu han cdi chung, nhung

cdi chung mang bdn chdt sdu sdc han cdi rieng

(cdi rieng: chi mdt sy vdt (SV), hien tugng (HT),

mdt qua trinh nhdt djnh; cdi chung: chi nhung

mat, nhung thudc tinh gidng nhau dugc lap

Igi trong cdc SV, HT vd cdc qua trinh rieng le).

Vi cdi chung chi ton tgi trong cdi rieng nen

mudn nhdn thuc dugc cdi chung, can phdi di

tu nhung cdi rieng, xudt phdt tu cdi rieng. Trong

DH todn, khi van dung cap phgm tru «cdi

chung - cdi rieng", chung ta khdng dugc tuyet

ddi hda mdt mat ndo.

Vi dy: Cho AABC vdi G Id trgng tdm, M Id

mdt diem bdt ki, chung minh rdng:

MA + MB+ \1C = JJIG.

Huang dan:

Ta cd: m+m+\^=(J^+cM)+0^+G^+0^+GQ

= 1AIG + (GA + CB+ GO = TJIG + 0 = iJlG.

Qua vi du tren, de nhdn thdy:

GA + GB + GC = 6 (1') Id «cd/ rieng" trong «cdi

chung" MA + MB + MC =3MG (1). Khi M -> G

thi MG^O vd XTA + JIB + JJC -> cZ4 + GB + GC{['\)

vd (1 ) cd mdi quan he vdi nhau, cd the dua

ddng thuc (1 ) ve (1) bdng cdch cdng hai ve

cua (V) vdi 3MG).

2) Khd ndng vd hien thyc: C g p p hgm tru

ndy cd mdi quan he bien chung, khdng tdch rdi

nhau vd cd the chuyen hda Idn nhau. Hien thyc

ludn dugc chudn bj bang khd ndng, con khd

ndng ludn hudng tdi bien thdnh hien thyc (kha

ndng: chi cdi se cd khi d mdt dieu kien thich

hgp; hien thyc: chi cdi dang cd, dang ton tgi).

Trong DH todn, mudn dy bdo khd ndng bien

ddi cua ddi tugng todn hgc, phdi nhdn thuc

dugc quy ludt cua dd'i tugng d'y. De hien thyc

hda khd ndng, can cd nhung tdc ddng phu

hgp vdi yeu cdu cua quy ludt, dap ung yeu

cdu thyc tiln.

Vidy: Cho phuang trinh (PT): ax2

+ by2

= 1 (1).

i) Vdi dieu kien ndo cua a vd b thi PT (1) Id

PT cua mdt dudng trdn?

ii) Vdi dieu kien ndo cua a vd b thi PT (1) Id

PT chinh tdc cua elip? Cua hypecbol?

Hudng ddn:

i) Khi a = b, a > 0 thi (1) Id PT cua mdt

dudng trdn: x2

+ y2

= —.

a

iij Khi 0 < a < b thi (1) Id PT chinh tdc cua

i. x2

v1

,

el'p: T + T = '

a b

Khi a > 0, b < 0 thi (1) Id PT chinh tdc cua

hypecbol: -—^- = 1.

* Tnrclng Dai hoc Tay Nguyen

** Throng Cao dang nghe thanh nien dan toe Tay Nguyen

Tap chi Giao due so 26 0 (k. 2.4/2011)