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27 Kvadratlar ayirmasining formulasi(3 soat)

Author:
October 21st, 2022

7-sinf, algebra.
(3 soatlik bayonnoma)
«_____» -____________, 20___-yil.
Mavzu: Kvadratlar ayirmasining formulasi.
1. Darsning maqsadi: umumiy ko‘paytuvchini qavsdan tashqariga chiqarish, guruhlash usuli,
yig‘indining va ayirmaning kvadrati, kvadratlar ayirmasining formulasi kabi mavzular yuzasidan
tushunchalar hosil qilish, o‘quvchilarning BKM larini shakllantirish, rivojlantirish va
mustahkamlash.
2. Darsning usuli: savol-javob, misol va masalalar yechish.
3. Darsning jihozi: DTS ko‘rgazmalari, tarqatma materiallar, kichik testlar.
DARSNING BORISHI:
1. Tashkiliy qism: o‘quvchilar bilan salomlashish, tozalikni tekshirish, davomatni aniqlash,
o‘quvchilarning dars mashg‘ulotlariga ruhiy jihatda tayyorliklarini aniqlash.
2. O‘tilgan mavzuni so‘rab baholash:
Savol: Ikki son yig‘indisining kvadrati nimaga teng?
Javob: Ikki son yig‘indisining kvadrati birinchi son kvadrati, qo‘shuv birinchi son bilan ikkinchi
son ko‘paytmasining ikkilangani, qo‘shuv ikkinchi sonning kvadratiga teng.
Savol: Ikki son ayirmasining kvadrati nimaga teng?
Javob: Ikki son ayirmasining kvadrati birinchi son kvadrati, ayiruv birinchi son bilan ikkinchi son
ko‘paytmasining ikkilangani, qo‘shuv ikkinchi sonning kvadratiga teng.
Savol: Ko‘phad birhadga qanday ko‘paytiriladi?
Javob: Ko‘phadni birhadga ko‘paytirish uchun ko‘phadning har bir hadini birhadga ko‘paytirish
va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad ko‘phadga qanday ko‘paytiriladi?
Javob: Ko‘phadni ko‘phadga ko‘paytirish uchun birinchi ko‘phadning har bir hadini ikkinchi
ko‘phadning har bir hadiga ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
Savol: Ko‘phad birhadga qanday bo‘linadi?
Javob: Ko‘phadni birhadga bo‘lish uchun ko‘phadning har bir hadini shu birhadga bo‘lish va
hosil bo‘lgan natijalarni qo‘shish kerak.
3. O‘tilgan mavzuni mustahkamlasi: o‘quvchilar tushunmagan savollar va ularga tushunarsiz
bo‘lgan jumlalar aniq va hayotiy misollar yordamida tushunturilib beriladi.
4. Yangi mavzuning bayoni:
Ikki son yig‘indisini ularning ayirmasiga ko‘paytiraylik:
(𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑎𝑏 + 𝑎𝑏 − 𝑏 2 = 𝑎2 − 𝑏 2
Demak, buni qisqacha quyidagicha yozish mumkin:
(𝒂 + 𝒃)(𝒂 − 𝒃) = 𝒂𝟐 − 𝒃𝟐 (𝟏)
Hosil bo‘lgan bu formula qisqa ko‘paytirish formulalaridan biri hisoblanadi hamda bu formula
hisoblashlarni soddalashtirish uchun qo‘llaniladi.
Masalan:
1) (2 + 𝑐)(2 − 𝑐) = 22 − 2𝑐 + 2𝑐 − 𝑐 2 = 4 − 𝑐 2 ;
2) (𝑥 − 5)(𝑥 + 5) = 𝑥 2 + 5𝑥 − 5𝑥 − 52 = 𝑥 2 − 25;
3) (2𝑥 + 𝑎)(2𝑥 − 𝑎) = (2𝑥)2 − 2𝑎𝑥 + 2𝑎𝑥 − 𝑎2 = 4𝑥 2 − 𝑎2 va h.k.
Yuqoridagi formuladan foydalangan holda, bu hisoblashlarni sodda shaklda quyidagicha ifodalash
mumkin:
1) (2 + 𝑐)(2 − 𝑐) = 4 − 𝑐 2 ; 2) (𝑥 − 5)(𝑥 + 5) = 𝑥 2 − 25; 3) (2𝑥 + 𝑎)(2𝑥 − 𝑎) = 4𝑥 2 − 𝑎2 .
Qoida: ikki son kvadratining ayirmasi shu sonlar ayirmasi bilan yig‘indisining ko‘paytmasiga
teng.
𝒂𝟐 − 𝒃𝟐 = (𝒂 − 𝒃)(𝒂 + 𝒃) (𝟏) Bu tenglik kvadratlar ayirmasining formulasi deb ataladi
Isboti: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑎𝑏 + 𝑎𝑏 − 𝑏 2 = 𝑎2 − 𝑏 2 .
(𝒂 + 𝒃)(𝒂 − 𝒃) = 𝒂𝟐 − 𝒃𝟐 (𝟐)
Bu tenglik qisqa ko‘paytirish formulalaridan biri hisoblanadi hamda undan ko‘phadlarni
ko‘paytuvchilarga ajratishda qo‘llaniladi.
Masalan:
1) 𝑎2 − 9 = (𝑎 − 3)(𝑎 + 3);
2) 9𝑥 2 − 4 = (3𝑥)2 − 22 = (3𝑥 − 2)(3𝑥 + 2);
3) (𝑎 − 𝑏)2 − 1 = (𝑎 − 𝑏 − 1)(𝑎 − 𝑏 + 1);
4) (𝑎 + 𝑏)2 − (𝑎 − 𝑐)2 = (𝑎 + 𝑏 − 𝑎 + 𝑐)(𝑎 + 𝑏 + 𝑎 − 𝑐) = (𝑏 + 𝑐)(2𝑎 + 𝑏 − 𝑐);
5) 25 ∙ 15 = (20 + 5)(20 − 5) = 400 − 25 = 375.
Misol va masalalar yechish:
386-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (𝑐 + 𝑑)(𝑐 − 𝑑) = 𝑐 2 − 𝑑2 ;
3) (𝑎 + 𝑐)(𝑐 − 𝑎) = (𝑐 + 𝑎)(𝑐 − 𝑎) = 𝑐 2 − 𝑎2 ;
2) (𝑝 + 𝑞)(𝑝 − 𝑞) = 𝑝2 − 𝑞 2 ;
4) (𝑚 − 𝑛)(𝑚 + 𝑛) = 𝑚2 − 𝑛2 .
387-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (𝑥 + 5)(𝑥 − 5) = 𝑥 2 − 25;
3) (𝑎 − 4)(4 + 𝑎) = (𝑎 − 4)(𝑎 + 4) = 𝑎2 − 16;
2) (𝑎 + 3)(𝑎 − 3) = 𝑎2 − 9;
4) (7 + 𝑥)(𝑥 − 7) = (𝑥 + 7)(𝑥 − 7) = 𝑥 2 − 49
388-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (2𝑏 + 𝑎)(2𝑏 − 𝑎) = (2𝑏)2 − (𝑎2 ) = 4𝑏 2 − 𝑎2 ;
2) (𝑐 + 3𝑑)(𝑐 − 3𝑑) = 𝑐 2 − (3𝑑)2 = 𝑐 2 − 9𝑑2 ;
3) (𝑦 + 6𝑥)(6𝑥 − 𝑦) = (6𝑥 + 𝑦)(6𝑥 − 𝑦) = (6𝑥)2 − (𝑦 2 ) = 36𝑥 2 − 𝑦 2 ;
4) (3𝑚 − 2𝑛)(2𝑛 + 3𝑚) = (3𝑚 − 2𝑛)(3𝑚 + 2𝑛) = (3𝑚)2 − (2𝑛)2 = 9𝑚2 − 4𝑛2.
389-misol. (𝑎 − 𝑏)(𝑎 + 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1
1
1
1 2
1
1
1) (4𝑑 − 2) (2 + 4𝑑) = (4𝑑 − 2) (4𝑑 + 2) = (4𝑑)2 − (2) = 16𝑑2 − 4;
5
5
5
5
2
5
25
2) (6 𝑎 − 𝑏) (𝑏 + 6 𝑎) = (6 𝑎 − 𝑏) (6 𝑎 + 𝑏) = (6 𝑎) − (𝑏 2 ) = 36 𝑎2 − 𝑏 2;
1
1
1
1
1
2
1
2
1
1
2
3
2
3
2
2
3
2
4
9
3) (2 𝑦 − 3 𝑥) (2 𝑦 + 3 𝑥) = (2 𝑦) − (3 𝑥) = 4 𝑦 2 − 9 𝑥 2 ;
4) (3 𝑚 + 4 𝑛) (3 𝑚 − 4 𝑛) = (3 𝑚) − (4 𝑛) = 9 𝑚2 − 16 𝑛2 .
391-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (3𝑎2 + 4𝑏 2 )(3𝑎2 − 4𝑏 2 ) = (3𝑎2 )2 − (4𝑏 2 )2 = 9𝑎4 − 16𝑏 4;
2) (2𝑚4 − 5𝑛2 )(5𝑛2 + 2𝑚4 ) = (2𝑚4 − 5𝑛2 )(2𝑚4 + 5𝑛2 ) = (2𝑚4 )2 − (5𝑛2 )2 = 2𝑚8 − 5𝑛4;
3) (0,2𝑡 3 + 0,5𝑝4 )(0,5𝑝4 − 0,2𝑡 3 ) = (0,5𝑝4 + 0,2𝑡 3 )(0,5𝑝4 − 0,2𝑡 3 ) = (0,5𝑝4 )2 − (0,2𝑡 3 )2 = 0,25𝑝8 − 0,4𝑡 6 ;
4) (1,2𝑎2 − 0,3𝑏 2 )(1,2𝑎2 + 0,3𝑏2 ) = (1,2𝑎2 )2 − (0,3𝑏2 )2 = 1,44𝑎4 − 0,9𝑏4 .
392-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
3
1
1
3
3
2
4
2
4
2
1
3
1
2
3
1
2
9
1
1) (4 𝑎2 − 2 𝑏 3 ) (2 𝑏 3 + 4 𝑎2 ) = (4 𝑎2 − 2 𝑏 3 ) (4 𝑎2 + 2 𝑏 3 ) = (4 𝑎2 ) − (2 𝑏 3 ) = 16 𝑎4 − 4 𝑏 6 ;
2
2
4
4
16
2) (3 𝑥 4 − 5 𝑦 5 ) (3 𝑥 4 − 5 𝑦 5 ) = (3 𝑥 4 ) − (5 𝑦 5 ) = 9 𝑥 8 − 25 𝑦10 ;
1
1
1
2
1
3) (0,5𝑞 + 3 𝑝2 ) (0,5𝑞 − 3 𝑝2 ) = (0,5𝑞)2 − (3 𝑝2 ) = 0,25𝑞2 + 9 𝑝4 ;
3
4
3
4
3
4
3
4
3
4
2
4) (1,5𝑐 2 − 𝑏) ( 𝑏 + 1,5𝑐 2 ) = (1,5𝑐 2 − 𝑏) (1,5𝑐 2 + 𝑏) = (1,5𝑐 2 )2 − ( 𝑏) = 2,25𝑐 4 −
9 2
𝑏 .
16
393-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (3𝑥 2 𝑦 − 4𝑥𝑦 2 )(3𝑥 2 𝑦 + 4𝑥𝑦 2 ) = (3𝑥 2 𝑦)2 − (4𝑥𝑦 2 )2 = 9𝑥 4 𝑦 2 − 16𝑥 2 𝑦 4 ;
2) (5𝑎𝑏 2 + 2𝑎2 𝑏)(5𝑎𝑏 2 − 2𝑎2 𝑏) = (5𝑎𝑏 2 )2 − (2𝑎2 𝑏)2 = 25𝑎2 𝑏4 − 4𝑎4 𝑏2 ;
3) (7𝑎𝑏 + 𝑥 2 𝑦 3 )(7𝑎𝑏 − 𝑥 2 𝑦 3 ) = (7𝑎𝑏)2 − (𝑥 2 𝑦 3 )2 = 49𝑎2 𝑏 2 − 𝑥 4 𝑦 6 ;
4) (𝑎𝑏 3 − 4𝑥𝑦)(𝑎𝑏 3 + 4𝑥𝑦) = (𝑎𝑏 3 )2 − (4𝑥𝑦)2 = 𝑎2 𝑏 6 − 16𝑥 2 𝑦 2.
394-misol. (𝑎 − 𝑏)(𝑎 + 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (3 + 𝑥)(3 − 𝑥)(9 + 𝑥 2 ) = (9 − 𝑥 2 )(9 + 𝑥 2 ) = 81 − 𝑥 4 ;
2) (𝑥 2 + 1)(𝑥 + 1)(𝑥 − 1) = (𝑥 2 + 1)(𝑥 2 − 1) = 𝑥 4 − 1;
3) (4𝑥 2 + 𝑦 2 )(2𝑥 + 𝑦)(2𝑥 − 𝑦) = (4𝑥 2 + 𝑦 2 )(4𝑥 2 − 𝑦 2 ) = 16𝑥 4 − 𝑦 4 ;
4) (3𝑎 − 2𝑏)(3𝑎 + 2𝑏)(9𝑎2 + 4𝑏 2 ) = (9𝑎2 − 4𝑏 2 )(9𝑎2 + 4𝑏 2 ) = 81𝑎4 − 16𝑏 4.
396-misol. Qisqa ko‘paytirish formulalaridan foydalanib hisoblang:
1) 27 ∙ 33 = (30 − 3)(30 + 3) = 900 − 9 = 891;
2) 44 ∙ 36 = (40 + 4)(40 − 4) = 1 600 − 16 = 1 584;
3) 84 ∙ 76 = (80 + 4)(80 − 4) = 6 400 − 16 = 6 384;
4) 201 ∙ 199 = (200 + 1)(200 − 1) = 40 000 − 1 = 39 999.
397-misol. Soddalashtiring:
1) (𝑐 − 3)2 − (𝑐 + 3)(3 − 𝑐) = 𝑐 2 − 6𝑐 + 9 − (9 − 𝑐 2 ) = 𝑐 2 − 6𝑐 + 9 − 9 + 𝑐 2 = 2𝑐 2 − 6𝑐;
2) (𝑎 + 2)2 − (𝑎 + 2)(2 − 𝑎) = 𝑎2 + 4𝑎 + 4 − (4 − 𝑎2 ) = 𝑎2 + 4𝑎 + 4 − 4 + 𝑎2 = 2𝑎2 + 4𝑎;
3) (2𝑥 + 3𝑦)(2𝑥 − 3𝑦) + (2𝑥 + 3𝑦)2 = 4𝑥 2 − 9𝑦 2 + 4𝑥 2 + 12𝑥𝑦 + 9𝑥 2 = 8𝑥 2 + 12𝑥𝑦;
4) (3𝑎 − 4𝑏)(3𝑎 + 4𝑏) − (3𝑎 − 4𝑏)2 = 9𝑎2 − 16𝑏 2 − 9𝑎2 + 24𝑎𝑏 − 16𝑏 2 = 24𝑎𝑏 − 32𝑏2 ;
5)(−𝑏 − 𝑎)(𝑎 + 𝑏) + 𝑎2 + 𝑏 2 = −𝑎𝑏 − 𝑏 2 − 𝑎2 − 𝑎𝑏 + 𝑎2 + 𝑏 2 = −2𝑎𝑏;
6) (𝑏 − 𝑎)(−𝑎 − 𝑏) + 2𝑏 2 = −𝑎𝑏 − 𝑏 2 + 𝑎2 + 𝑎𝑏 + 2𝑏 2 = 𝑎2 + 𝑏 2 .
398-misol. Ifodaning qiymatini toping:
1) 4𝑚 − (𝑚 + 3)2 + (𝑚 − 3)(𝑚 + 3), bunda 𝑚 = −2,4;
𝑎) 4𝑚 − (𝑚 + 3)2 + (𝑚 − 3)(𝑚 + 3) = 4𝑚 − 𝑚2 − 6𝑚 − 9 + 𝑚2 − 9 = −2𝑚;
𝑏) − 2𝑚 = −2 ∙ (−2,4) = 4,8;
2) (3𝑥 + 4)2 − 10𝑥 − (𝑥 − 4)(4 + 𝑥), bunda 𝑥 = −0,1;
𝑎) (3𝑥 + 4)2 − 10𝑥 − (𝑥 − 4)(4 + 𝑥) = 9𝑥 2 + 24𝑥 + 16 − 10𝑥 − 𝑥 2 + 16 = 8𝑥 2 + 14𝑥 + 32;
𝑏) 8𝑥 2 + 14𝑥 + 32 = 8 ∙ (−0,1)2 + 14 ∙ (−0,1) + 32 = 8 ∙ 0,01 − 1,4 + 32 = 0,08 − 1,4 + 32 = 30,68;
1
3) 2(𝑘 − 7)(𝑘 + 5) − (𝑘 − 5)2 − (𝑘 − 7)(7 + 𝑘), bunda 𝑘 = − 2 = −0,5;
𝑎) 2(𝑘 − 7)(𝑘 + 5) − (𝑘 − 5)2 − (𝑘 − 7)(7 + 𝑘) = 2(𝑘 2 + 5𝑘 − 7𝑘 − 35) − (𝑘 2 − 10𝑘 + 25) −
−(𝑘 2 − 49) = 2𝑘 2 − 4𝑘 − 70 − 𝑘 2 + 10𝑘 − 25 − 𝑘 2 + 49 = 6𝑘 − 46;
𝑏) 6𝑘 − 46 = 6 ∙ (−0,5) − 46 = −3 − 46 = −49.
1
4) (𝑎 + 3)2 + (𝑎 − 3)(3 + 𝑎) − 2(𝑎 + 2)(𝑎 − 4), bunda 𝑎 = − 5 = −0,2;
𝑎) (𝑎 + 3)2 + (𝑎 − 3)(3 + 𝑎) − 2(𝑎 + 2)(𝑎 − 4) = 𝑎2 + 6𝑎 + 9 + 𝑎2 − 9 − 2(𝑎2 − 4𝑎 + 2𝑎 − 8) =
= 𝑎2 + 6𝑎 + 9 + 𝑎2 − 9 − 2𝑎2 + 4𝑎 + 16 = 10𝑎 + 16;
𝑏) 10𝑎 + 16 = 10 ∙ (−0,2) + 16 = −2 + 16 = 14.
399-misol. Tenglamani yeching:
1) (2𝑥 + 3)2 − 4(𝑥 − 1)(𝑥 + 1) = 49;
4𝑥 2 + 12𝑥 + 9 − 4(𝑥 2 + 1) = 49
4𝑥 2 + 12𝑥 + 9 − 4𝑥 2 − 4 = 49
12𝑥 = 49 − 5
12𝑥 = 44
44
𝑥 = 12
𝑥=
10
.
3
2) (3𝑥 + 4)2 − (3𝑥 − 1)(1 + 3𝑥) = 49;
9𝑥 2 + 24𝑥 + 16 − (9𝑥 2 − 1) = 49
9𝑥 2 + 24𝑥 + 16 − 9𝑥 2 + 1 = 49
24𝑥 = 49 − 17
24𝑥 = 32
32
𝑥 = 24
4
3
𝑥= .
3) 𝑥 3 + 2𝑥 2 − 9𝑥 − 18 = 0;
(𝑥 3 + 2𝑥 2 ) − (9𝑥 + 18) = 0
𝑥 2 (𝑥 + 2) − 9(𝑥 + 2) = 0
(𝑥 2 − 9)(𝑥 + 2) = 0
1) 𝑥 2 − 9 = 0; 2) 𝑥 + 2 = 0.
𝑥2 = 9
𝑥 =0−2
𝑥 = √9
𝑥 = −2 .
𝑥 = ±3 .
Javob: 𝑥 = ±3 va 𝑥 = −2.
4) 𝑦 3 − 3𝑦 2 − 4𝑦 + 12 = 0;
(𝑦 3 − 3𝑦 2 ) − (4𝑦 − 12) = 0
𝑦 2 (𝑦 − 3) − 4(𝑦 − 3) = 0
(𝑦 2 − 4)(𝑦 − 3) = 0
1) 𝑦 2 − 4 = 0; 2) 𝑦 − 3 = 0.
𝑦2 = 4
𝑦 =0+3
𝑦 = √4
𝑦=3 .
𝑦 = ±2.
Javob: 𝑦 = ±2 va 𝑦 = 3.
400-misol. Kvadratning ikki qarama-qarshi tomonining har biri 8 𝑠𝑚 ga uzaytirildi, qolgan ikki
tomoni esa shuncha qisqartirildi. Shaklning yuzi qanday o‘zgardi?
Yechilishi:
𝑆 = (𝑥 + 8)(𝑥 − 8);
𝑆 = 𝑥 2 − 64;
Javob: 64 𝑠𝑚2 ga kamaygan.
5. O‘tilgan mavzuni mustahkamlash: o‘quvchilar tushunmagan savollarni aniq misollar yordamida
tushuntiraman.
6. O‘tilgan mavzuni so‘rab baholash:
Savol: Ikki son yig‘indisining kvadrati nimaga teng?
Javob: Ikki son yig‘indisining kvadrati birinchi son kvadrati, qo‘shuv birinchi son bilan ikkinchi
son ko‘paytmasining ikkilangani, qo‘shuv ikkinchi sonning kvadratiga teng.
Savol: Ikki son ayirmasining kvadrati nimaga teng?
Javob: Ikki son ayirmasining kvadrati birinchi son kvadrati, ayiruv birinchi son bilan ikkinchi son
ko‘paytmasining ikkilangani, qo‘shuv ikkinchi sonning kvadratiga teng.
Savol: Kvadratlar ayirmasi nimaga teng?
Javob: Ikki son kvadratining ayirmasi shu sonlar ayirmasi bilan yig‘indisining ko‘paytmasiga
teng.
Savol: Ko‘phad ko‘phadga qanday ko‘paytiriladi?
Javob: Ko‘phadni ko‘phadga ko‘paytirish uchun birinchi ko‘phadning har bir hadini ikkinchi
ko‘phadning har bir hadiga ko‘paytirish va hosil bo‘lgan ko‘paytmalarni qo‘shish kerak.
7. Uyga vazifa: O‘tilgan mavzuni o‘qib o‘rganish va misollar yechish.
390-misol. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 formuladan foydalanib, ko‘paytirishni bajaring:
1) (𝑐 2 + 𝑑2 )(𝑐 2 − 𝑑2 ) = (𝑐 2 )2 − (𝑑2 )2 = 𝑐 4 − 𝑑4 ;
2) (𝑎2 + 𝑏 3 )(𝑎2 − 𝑏 3 ) = (𝑎2 )2 − (𝑏 3 )2 = 𝑎4 − 𝑏 6;
3) (𝑥 4 − 𝑦 3 )(𝑦 3 + 𝑥 4 ) = (𝑥 4 − 𝑦 3 )(𝑥 4 + 𝑦 3 ) = (𝑥 4 )2 − (𝑦 3 )2 = 𝑥 8 − 𝑦 6 ;
4) (𝑚3 − 𝑛3 )(𝑚3 + 𝑛3 ) = (𝑚3 )2 − (𝑛3 )2 = 𝑚6 − 𝑛6.
395-misol. Qisqa ko‘paytirish formulalaridan foydalanib hisoblang:
1) 48 ∙ 52 = (50 − 2)(50 + 2) = 2 500 − 4 = 2 496;
2) 68 ∙ 72 = (70 − 2)(70 + 2) = 4 900 − 4 = 4 896;
3) 43 ∙ 37 = (40 + 3)(40 − 3) = 1 600 − 9 = 1 591;
4) 47 ∙ 53 = (50 − 3)(50 + 3) = 2 500 − 9 = 2 491.
401-misol. Hisoblang:
Yechilishi:
54 ∙0,128−53 ∙0,628∙5
125∙0,25
=
625∙0,128−125∙3,14
125∙0,25
=
125∙(5∙0,128−3,14)
125∙0,25
=
0,64−3,14
0,25
=
−2,5
0,25
= −10.
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