Kirish i-bob. Matematika fanlarini o’qitishda zamonaviy axborot texnologiyalaridan foydalanish metodikasi

2-misol. Ushbu



ifodadan butun qism ajratamiz. Buning uchun suratdagi ko’phadni maxrajdagi ko’phadga bo’lish lozim. Bo’lishni “burchakli bo’lish” usulida bajaramiz:

3x^4-10ax^3+22a^2x^2-24a^3x+10a^4	x^2-2ax+3a^2
3x^4-6ax^3+9a^2x^2	3x^2-4ax+5a^2
-4ax^3+13a^2x^2-24a^3x
-4ax^3+8a^2x^2-12a^3x
5a^2x^2-12a^3x+10a^4
5a^2x^2-10a^3x+15a^4
-2a^3x-5a^4

Demak,

Aytaylik, φ(x)=b₀+b₁x+b₂x²+…+b_nxⁿ ko`phad berilgan bo`lsin . Darajasi n ga teng va bosh koeffisenti b _n 0 bo`lgan har qanday φ(x) ko`phadning bosh koeffisentini doimo 1 ga keltirib olish mumkin . Buning uchun ko`phadni qarash kifoya . g(x) ko`phadddan boshqa bosh koeffisenti ixtiyoriy bo`lgan m≥n darajali f(x)=α<sub>0+</sub> α<sub>1</sub>x+ α<sub>2</sub>x<sup>2</sup>+ …+ α<sub>m</sub>x<sup>m</sup> ko`phad berilgan bo`lsin .

Agar f(x) ko`phad n- darajali ko`phad bo`lsa u deg f(x) =n kabi yoziladi.
Teorema. Har qanday f(x) va g(x) 0 ko`phadlar uchun shunday yagona h(x) va r(x) ko`phadlar mavjudki , ular uchun deg r(x) < degg(x) va deg h(x)f(x)=g(x) h(x)+ r(x) (2.2)
Isboti Agar f(x) ko`phaddan a _mx^m-ng(x) ko`phaddi ayirsak , f(x)- a <sub>m</sub>x<sup>m-n</sup>g(x) = r<sub>1</sub>(x) ko`phadda a _mx^m had bo`lmaydi . Bu yerda quyidagi ikki hol bo`lishi mumkin :
a) r₁(x) x ning darajasi g(x) ning darajasidan kichik ;
b) r₁(x) ning rarajasi g(x) darajasidan katta yoki unga teng .
Agar a) hol yuz bersa , h(x) =a _mx^m-n ; r(x)= r₁(x) bo`lib , teorema isbotlangan bo`ladi . Biz b) hol ustida to`xtalib o`tamiz . faraz qilaylik , dar r₁(x) ≥ deg g(x) bo`lib , r<sub>1</sub>(x) = c<sub>0</sub>+c<sub>1</sub>x +c<sub>2</sub>x<sup>2</sup>+… +c<sub>k</sub>x<sup>k</sup> ko`rinishga ega bo`lsin .
Endi g(x) ko`phadni c_kx^k-n ga ko`paytirib , natijasini r<sub>1</sub>(x) dan ayiramiz . U holda r<sub>1</sub>(x) - c<sub>k</sub>x<sup>k-n</sup>g(x) = r<sub>2</sub>(x) bo`lib , r₂(x) ko`phadga c<sub>k</sub>x<sup>k</sup> had bo`lmaydi .
r₂(x)=d₀+d₁x+d₂x²+…+d_lx^l bo`lsin . Bu yerda yana yuqoridagi ikki holdan biri yuz berishi mumkin :
1) Agar l≥n bo`lsa , ushbu ayrmani tuzamiz :
r₂(x)- d₁x^l-n g(x)= r₃(x) ,
jarayonni davom ettirib , biror v qadamdan so`ng dar r<sub>v</sub>(x)< dar g(x) ga erishamiz .Boshqacha aytganda , r<sub>v-1</sub>(x)- g(x)=r<sub>v</sub>(x) tenglikda dar r<sub>v</sub>(x)< dar g(x) bo`ladi .
Endi ushbu tengliklarni hadlab qo`shamiz :
f(x)- a <sub>m</sub>x<sup>m-n</sup>g(x)= r<sub>1</sub>(x)
r<sub>1</sub>(x)- c <sub>k</sub>x<sup>k-n</sup> g(x)= r<sub>2</sub>(x),
r<sub>2</sub>(x)- d <sub>l</sub>x<sup>l-n</sup> g(x)= r<sub>3</sub>(x),
……………………….
r<sub>v-1</sub>(x)- g(x)= r<sub>v</sub>(x).
unda
f(x)-( a <sub>m</sub>x<sup>m-n</sup>+ c <sub>k</sub>x<sup>k-n</sup> + d <sub>l</sub>x<sup>l-n</sup>+…+ )g(x)= r<sub>v</sub>(x)
hosil bo`ladi . bu yerda
a _mx^m-n+ c _kx^k-n +…+ =h(x) va r_v(x) =r(x)
desak, f(x)=g(x)·h(x)+r(x) tenglik hosil bo`ladi .
f(x)=g(x)·h(x)+r(x) tenglikdagi f(x) bo`linuvchi , g(x) bo`luvchi , h(x) chala bo`linma , r(x) esa qoldiq ko`phadlar deyiladi .
Endi (1) tenglikning yagonaligini isbotlaymiz .
Aytaylik , (1) shartni qanoatlantiruvchi yana bir juft va
Ko`phad mavjud , yani
f(x)=g(x) · + (2.3)
tenglik o`rinli bo`lsin . (2.2) va (2.3) tengliklarni hadlab ayirib
o=g(x)(h(x)- )+(r(x)- )
yoki
g(x)·(h(x)- )= - r(x) (2.4)
ni hosil qilamiz . Bu yerda r(x) va ning aniqlanishiga asosan dar
( -r(x))< dar g(x) bo`ladi . Agar chap tomonda h(x)- 0 bo`lsa , ( -r(x)) ning darajasi (2.4) ga asosan g(x) ning darajasidan kichik emas. Bu esa r(x) va ning aniqlanishiga ziddir . Shuning uchun h(x) = bo`ladi . bunga ko`ra (3) dan r(x)= kelib chiqadi .
Bu teoremani bazan f(x) ko`phadni g(x) ko`phadga qoldiqli bo`lish teoremasi deb yuritiladi .