Sinf to’garak a’zolari Matematika Mavzu: Natural son tushunchasi va ular ustida amallar. Darsdan maqsad

IV. Yangi mavzuni mustahkamlash: mavzuga doir darslikdagi tegishli misolni yechish
V. Darsni yakunlash va baholash: darsda ishtirok etgan o’quvchilar baholanadi, ishtirok etmaganlari esa ogohlantiriladi.
VI. Uyga vazifa: -mavzu -misol - bet
Ko’rildi O’IBDO’: .


-sinf to’garak a’zolari Matematika
Mavzu:Qisqa ko’paytirish formulalari.
Darsdan maqsad.
a)Ta’limiy:–O’quvchilarga Yig’indining kvadrati haqida tushunchalar berish
b) Tarbiyaviy: - O’quvchilarni matematika faniga qiziqishini oshirish.
c) Rivojlantiruvchi: - Mavzuga doir misollar yechish.
Dars turi: Darslik bilan ishlash Dars mtodi: an’anaviy
Dasr jihozi: matematikadan misollar to’plami va 5-sinf Matematika o’quv qo’llanmasi
Dars tafsilotlari

1	Tashkiliy qism	3 daqiqa
2	O’tilgan mavzuni so’rash	6 daqiqa
3	Yangi mavzu bayoni	27 daqiqa
4	Yangi mavzuni mustahkamlash	5 daqiqa
5	Darsni yakunlash va baholash	2 daqiqa
6	Uyga vazifa	daqiqa

I. Tashkiliy qism a) Salomlashish b) Davomatni aniqlash d) O’quvchilarning darsga tayyorgarligini tekshirish
II.O’tilgan mavzuni so’rash: o’quvchilar o’rtasida savol-javob o’yinini tashkil qilish.
III.Yangi mavzu bayoni
Yangi bilimni o’zlashtirish
Ikki son yig'indisining kvadrati birinchi son kvadrati, qo'shuv birinchi son bilan ikkinchi son ko'paytmasining ikkilangani, qo'shuv ikkinchi son kvadratiga teng.
Ikkita son yig'indisining kvadrati (a + b)² ni keltirib chiqaramiz. Ko'phadni ko'phadga ko'paytirish qoidasidan foydalanib quyidagini hosil qilamiz:
(a + b)² = (a + b)(a + b) = a² + ab + ab + b² = a² + 2ab + b² .
Demak, (a + b)² = a² + 2ab + b² .
Masalan: 1) 99² = (100 – 1)² = 100 ² – 2 · 100 · 1 + 1² = 10000 – 200 + 1 = 9801 .
2) 73² = (70 + 3)² = 70 ² + 2 · 70 · 3 + 3² = 4900 + 420 + 9 = 5329 .
(a + b)³ = (a + b) (a + b)² = (a + b)(a² + 2ab + b²) =
a³ + 2a²b + ab² + a²b + 2ab² + b³ = a³ + 3a²b + 3ab² + b³ .
Demak, (a + b)³ = a³ + 3a²b + 3ab² + b³ .
1 – misol.
1) (c + d)² = (c + d)(c + d) = c² + cd + cd + d² = c² + 2cd + d²
2) (2 + x)² = (2 + x)(2 +x) = 4 + 2x + 2x + x² = 4 + 4x + x²
3) (y +3d)² = (y + 3)(y + 3) = y² + 3y +3y + 9 = y² + 6y + 9
4) (c + d)² = (c + d)(c + d) = c² + cd + cd + d² = c² + 2cd + d²
Ikkita son ayirmasining kvadrati (a – b)² ni keltirib chiqaramiz. Ko'phadni ko'phadga ko'paytirish qoidasidan foydalanib quyidagini hosil qilamiz:


(a – b)² = (a – b)(a – b) = a² – ab – ab + b² = a² – 2ab + b² .
Demak,
(a – b)² = a² – 2ab + b² .


Ikki son ayirmasining kvadrati birinchi son kvadrati, ayiruv birinchi son bilan ikkinchi son ko'paytmasining ikkilangani, qo'shuv ikkinchi son kvadratiga teng
Masalan:
1) 1,001² = (1 + 0,001)² ≈ 1 + 2 · 0,001 = 1 + 0,002 = 1,002 .
2) 0,998² = (1 – 0,002)² ≈ 1 – 2 · 0,002 = 1 – 0,004 = 0,996 .
3) a⁴ – 8a²b³ + 16b⁶ = (a²)² – 2 · a² · 4b³ + (4b³)² = (a² – 4b³)² .
(a – b)³ = (a – b) (a – b)² = (a – b)(a² – 2ab + b²) =
a³ – 2a²b + ab² – a²b + 2ab² – b³ = a³ – 3a²b + 3ab² – b³ .
Demak, (a – b)³ = a³ – 3a²b + 3ab² – b³ .
Hosil qilingan formulalar mos ravishda yig'indining kubi va ayirmaning kubi deb ataladi. Bu formulalar ham qisqa ko'paytirish formulalari hisoblanadi
Ikki son kvadratlari ayirmasi a² – b² ifoda berilgan bo‘lsin. Bu ifodaga ab birhadni qo‘shamiz va ayiramiz:
a² – b² = a² – b² + ab – ab .
Hosil bo‘lgan ko‘phadning 1- va 3- hadlarini guruhlaymiz, 2- va 4- hadlarini esa “–” ishorasi bilan guruhlaymiz, so‘ngra ko‘paytuvchilarga ajratamiz:
a² – b² + ab – ab = (a² + ab) – (b² + ab) = a(a + b) – b(b + a) =
= (a + b)(a – b) .
Demak, a² – b² = (a + b)(a – b) .
Ikki son kvadratlarining ayirmasi shu sonlar ayirmasi bilan ular yig'indisining ko'paytmasiga teng.
Masalan:1-misol
1) m²n²– 9k² = (mn + 3k) (mn – 3k) ;
2) (2a²b + 5ab²) (2a²b – 5ab²) = 4a⁴b² – 25a²b⁴ .
3) 37 · 43 = (40 – 3) · (40 + 3) = 40² – 3² = 1600 – 9 = 1591 ;
4) 102 · 98 = (100 + 2) · (100 – 2) = 100² – 2² = 10000 – 4 = 9996 .
5) a² – 16 = (a + 4)(a – 4) ;
6) 9b⁶ – 0,81c⁴ = (3b³)² – (0,9c²)² = (3b³ + 0,9c²)( 3b³ – 0,9c²) .
IV. Yangi mavzuni mustahkamlash: mavzuga doir darslikdagi tegishli misolni yechish
V. Darsni yakunlash va baholash: darsda ishtirok etgan o’quvchilar baholanadi, ishtirok etmaganlari esa ogohlantiriladi.
VI. Uyga vazifa: -mavzu -misol - bet
Ko’rildi O’IBDO’: .

-sinf to’garak a’zolari Matematika
Mavzu: 7- sinf Algebra kursini takrorlash
Darsdan maqsad.
O‘quvchilarni quyidagi tushunchalar bilan tanishtirish:
a) Ta’limiy:
– 7- sinf Algebra kursida o’rganilgan algebraik ifodalar, birhadlar, ko’phadlar, ko’phadni ko’paytuvchilarga ajratish usullarini takrorlash
b) Tarbiyaviy:
- O’quvchilarda vatanparvarlik ruhini shakllantirish.
c) Rivojlantiruvchi:
- Mavzuga doir misollar yechish.
Dars turi: Darslik bilan ishlash Dars mtodi: an’anaviy
Dasr jihozi: matematikadan misollar to’plami va 5-sinf Matematika o’quv qo’llanmasi
Dars tafsilotlari

1	Tashkiliy qism	3 daqiqa
2	O’tilgan mavzuni so’rash	6 daqiqa
3	Yangi mavzu bayoni	27 daqiqa
4	Yangi mavzuni mustahkamlash	5 daqiqa
5	Darsni yakunlash va baholash	2 daqiqa
6	Uyga vazifa	daqiqa

I. Tashkiliy qism a) Salomlashish b) Davomatni aniqlash d) O’quvchilarning darsga tayyorgarligini tekshirish
II.O’tilgan mavzuni so’rash: o’quvchilar o’rtasida savol-javob o’yinini tashkil qilish.
III.Yangi mavzu bayoni
1- misol
1) S = 2(ab +ac +bc), bunda a = 5, b = 4, c = 10;

2) , bunda h = 12, a = 10, b = 8’

3) , bunda a = 10, b = 40, n = 16;

4) , bunda a = 30, b = 20, h = 25;
2 – misol
1) 7a – (5a + 4b) = 7a - 5a - 4b = 2a - 4b
2) 9x – (7y +4x) = 9x – 7y – 4x = 9x – 4x - 7y = 5x – 7y
3) –(2a -3b) – (-a + 3b)=-2a + 3b + a – 3b = -2a + a = -a
4) 8x – (3y + 5x) – (-2y - x)= 8x - 3y - 5x + 2y + x = 4x – y
3– misol
1) ba 8ac, bunda , b = - 3, c = 2;

2) , bunda x = 3,



4 – misol
1) 1,2ab + 0,8 – 0,2ab + 2,2b ²+ 2ab= 1,2ab - 0,2ab + 2ab+
+ 0,8b² + 2,2b² = 3ab + 3b²
2) 3a²2a² + 3b²4a² - 2a²5b ²- 3a2a² - a³2a = 6a⁴ + 12a²b² - 10a²b² –
- 6a²b² -2a⁴ = 6a⁴-2a⁴ + 2a²b²= 4a⁴ + 2a²b²


5 –misol.
1) (x + y)(a - b) = ax –bx +ay –by
2) (a – b + c)(a - c) =a² – ac –ab+ bc +ac – c²
3) (a² – b²)(a+b) = a³ + a²b – ab² – b³
4) (a-3)(a-2) – (a-1)(a-4)= a² - 2a - 3a + 6 - a² + 5a-4 = 2
6 – misol
1) 5a² – 15a⁴ + 10a⁶ = 5a²(1 – 3a² + 2a³)
2) 9a³ + 12a² – 6a = 3a(3a² + 4a -2)
3) a(x+y) - b(x+y) = (a-b)(x+y)
4) (x-1) – a(x-1) = (1-a)(x-1)
5) 4(a-3) + a(3-a) = (4-a)(a-3)
6) a²(1-a) + 4a(a-1) = (a²-4a)(1-a)


7 – misol
1) ay + zy – 2ap -2zp= y(a+z) – 2p(a+z)=(y-2p)(a+z)
2) 5ac-6bd+5ad – 6bc = 5a(c+d)-6b(c+d)= (5a-6b)(c+d)
3) a(5a-4b) – 10a+8b = 5a² – 4ab -10a+8b = (5a-4b)(a-2)
4) 4ab-6cd-12ad+2bc= 4a(b-3d)- 2c(3a+b) = (4a+2c)(b-3d)
8 – misol
1)
2)
3)
4)