Таълим ва инновацион тадқиқотлар (2021 йил Махсус сон)
ISSN 2181-1709 (P)
22
Education and innovative research 2021 y. Sp.I.
1
elementlari mavzulariga bag
ʻ
ishlanadi. Mavzular sodda va batafsil yoritilgan. Misol va
masalalar esa kundalik turmush tarziga mos qilib o
ʻ
quvchi osonlik bilan tushunib bemalol mustaqil
tahlil qila oladi. Fan o
ʻ
qituvchisidan berilgan mavzularni sinfning har bir o
ʻ
quvchisiga samarali
yetkazib berishi uchun
barcha bilim va ko’nikmalarini ishga solib , zamonaviy ped texnologiyalar
va interfaol matodlardan foydalanish talab etiladi.
Biz quyida kitobning I bobida keltirilgan mavzularga doir
ba’zi misol va masalalarni
yechish usullarini tahlil qilib o
ʻ
tamiz.
1-misol.
𝑈𝑈 = {𝑥𝑥𝑥𝑥0 < 𝑥𝑥𝑥 ≤ 12, 𝑥𝑥𝑥 ∈ 𝑍𝑍},
𝐴𝐴 = {𝑥𝑥𝑥𝑥2 ≤ 𝑥𝑥𝑥 ≤ 7, 𝑥𝑥𝑥 ∈ 𝑍𝑍},
𝐵𝐵 = {𝑥𝑥𝑥𝑥3 ≤ 𝑥𝑥𝑥 ≤ 9, 𝑥𝑥𝑥 ∈ 𝑍𝑍},
𝐶𝐶 = {𝑥𝑥𝑥𝑥5 ≤ 𝑥𝑥𝑥 ≤ 7, 𝑥𝑥𝑥 ∈ 𝑍𝑍}
bo
ʻ
lsa, quyidagilarni toping: a)
𝐵𝐵´ ;
b)
𝐶𝐶´ ;
c)
𝐴𝐴´ ;
d)
𝐴𝐴 ∩ 𝐵𝐵 ;
e)
(𝐴𝐴 ∩ 𝐵𝐵)´ ;
f)
𝐴𝐴´ ∩ 𝐶𝐶
.
Yechish: Bu yerdagi
𝐴𝐴´
,
𝐵𝐵´
,
𝐶𝐶´
lar A, B, C to
ʻ
plamlarning to
ʻ
ldiruvchi to
ʻ
plamidir. To
ʻ
ldiruvchi
to
ʻplamning ta’rifini keltiramiz.
Ta’rif. A toʻ
plamning
𝐴𝐴´
to
ʻ
ldiruvchisi deb U universal to
ʻ
plamning A ga tegishli bo
ʻ
lmagan
barcha elementlari to
ʻ
plamiga aytiladi.
a)
𝐵𝐵´ = {1,2,10,11,12};
b)
𝐶𝐶´ = {1,2,3,4,8,9,10,11,12} ;
c)
𝐴𝐴´ = {1,8,9,10,11,12,13};
d)
𝐴𝐴 ∩ 𝐵𝐵 = {3,4,5,6,7} ;
e)
(𝐴𝐴 ∩ 𝐵𝐵)´ = {1,2,8,9,10,11,12};
f)
𝐴𝐴´ ∩ 𝐶𝐶 = {1,5,6,7,8,9,10,11,12}.
Ushbu misol yechimlarini Eyler-Venn diagrammalarida ko
ʻ
rsatish maqsadga muvofiq.
2-misol. U=N,
𝐴𝐴 = {6 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 31 𝑑𝑑𝑑𝑑𝑠𝑠 𝑘𝑘𝑠𝑠𝑘𝑘ℎ𝑠𝑠𝑘𝑘 𝑏𝑏𝑠𝑠
′
𝑙𝑙𝑠𝑠𝑑𝑑𝑠𝑠 𝑘𝑘𝑑𝑑𝑘𝑘𝑘𝑘𝑑𝑑𝑙𝑙𝑠𝑠𝑙𝑙𝑑𝑑𝑘𝑘𝑠𝑠},
𝐵𝐵 = {30 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑏𝑏𝑠𝑠′𝑙𝑙𝑙𝑙𝑙𝑙𝑘𝑘ℎ𝑠𝑠𝑙𝑙𝑑𝑑𝑘𝑘𝑠𝑠}
va
𝐶𝐶 = {30 𝑑𝑑𝑑𝑑𝑠𝑠 𝑘𝑘𝑠𝑠𝑘𝑘ℎ𝑠𝑠𝑘𝑘 𝑏𝑏𝑠𝑠
′
𝑙𝑙𝑠𝑠𝑑𝑑𝑠𝑠 𝑡𝑡𝑙𝑙𝑏𝑏 𝑠𝑠𝑠𝑠𝑠𝑠𝑙𝑙𝑑𝑑𝑘𝑘}
bo
ʻ
lsin. a) A, B, C
; b)
𝑠𝑠(𝐴𝐴 ∪ 𝐵𝐵 ∪ 𝐶𝐶) = 𝑠𝑠(𝐴𝐴) + 𝑠𝑠(𝐵𝐵) + 𝑠𝑠(𝐶𝐶) − 𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵) − 𝑠𝑠(𝐵𝐵 ∩ 𝐶𝐶) − 𝑠𝑠(𝐴𝐴 ∩ 𝐶𝐶) + 𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵 ∩
𝐶𝐶)
tenglikning bajarilishini tekshiring.
Yechish: a)
𝐴𝐴 = {6, 12,18,24,30}, 𝐵𝐵 = {1,2,3,5,6,10,15,30}
,
𝐶𝐶 = {2,3,5,7,11,13,17,19,23,29}
.
b)
n(A)- deb A to
ʻ
plam elementlari sonini belgilaymiz.
𝑠𝑠(𝐴𝐴) = 5 ,
𝑠𝑠(𝐵𝐵) = 8,
𝑠𝑠(𝐶𝐶) = 10,
𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵) = 2,
𝑠𝑠(𝐵𝐵 ∩ 𝐶𝐶) = 3,
𝑠𝑠(𝐴𝐴 ∩ 𝐶𝐶) = 0,
𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵 ∩ 𝐶𝐶) = 0
𝑠𝑠(𝐴𝐴 ∪ 𝐵𝐵 ∪ 𝐶𝐶) = 18.
𝑠𝑠(𝐴𝐴) + 𝑠𝑠(𝐵𝐵) + 𝑠𝑠(𝐶𝐶) − 𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵) − 𝑠𝑠(𝐵𝐵 ∩ 𝐶𝐶) − 𝑠𝑠(𝐴𝐴 ∩ 𝐶𝐶) + 𝑠𝑠(𝐴𝐴 ∩ 𝐵𝐵 ∩ 𝐶𝐶)
=
5+8+10-2-
3
-
0+0=18;
𝑠𝑠(𝐴𝐴 ∪ 𝐵𝐵 ∪ 𝐶𝐶) = 18
tenglik isbotlandi.
3.
Agar
𝐴𝐴 = {𝑥𝑥𝑥 ∈ 𝑁𝑁𝑁𝑥1 ≤ 𝑥𝑥𝑥 ≤ 3} , 𝐵𝐵 = {𝑦𝑦 ∈ 𝑁𝑁𝑁𝑥2 ≤ 𝑦𝑦 ≤ 4}
bo
ʻ
lsa,
𝐴𝐴 × 𝐵𝐵
ni toping va uni
dekart tekisligiga chizing.
𝐴𝐴 × 𝐵𝐵 = {(𝑥𝑥𝑥, 𝑦𝑦)𝑥1 ≤ 𝑥𝑥𝑥 ≤ 3 , 2 ≤ 𝑦𝑦 ≤ 4}
4.
Berilgan
𝑑𝑑 ∧ 𝑏𝑏
̅̅̅̅̅̅̅ ⇔ 𝑑𝑑̅ ∨ 𝑏𝑏̅
ifoda mantiq qonuni ekanligini isbotlang.
Yechish: Avvalo berilgan ifodaning rostlik jadvalini tuzamiz:
a
b
𝑑𝑑̅
𝑏𝑏̅
𝑑𝑑 ∧ 𝑏𝑏
𝑑𝑑 ∧ 𝑏𝑏
̅̅̅̅̅̅̅
𝑑𝑑̅ ∨ 𝑏𝑏̅
1
1
0
0
1
0
0
1
0
0
1
0
1
1
0
1
1
0
0
1
1
0
0
1
1
0
1
1
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