2. Mantiqiy masalalarni yechishning asosiy usullari

1-misol. Tenglamaning barcha natural 𝑥, 𝑦 yechimlarini toping
7𝑥 + 13𝑦 = 113.
Yechish: Tenglamadan 𝑥 ni topamiz va undan butun qismini ajratib yozamiz:
𝑥 =
113 − 13𝑦
7 7 7 7 7
= = − = 16 − 2𝑦 −
102 − 14𝑦 + 1 + 𝑦
102 − 14𝑦
1 + 𝑦
1 + 𝑦
Bu ifoda natural son bo‘lishi uchun ravshanki, 1+𝑦 ifoda butun son bo’lishi kerak.
7
Keyingi xulosalar
1 + 𝑦
7
= 𝑘, 𝑘Є𝑍 ; 1 + 𝑦 = 7𝑘; 𝑦 = 7𝑘 − 1;
𝑥 =
113 − 13(7𝑘 − 1) 126 − 91𝑘
7 7
=
= 18 − 13𝑘 Є 𝑁.
{18 − 13𝑘 > 0
7𝑘 − 1 > 0
dan 𝑘 = 1 bo‘lishi kelib chiqadi.
𝑘 = 1 da 𝑥 = 18 − 13 ∙ 1 = 5,
Javob: x=5, y=6.
𝑦 = 7 ∙ 1 − 1 = 6
2‐misol. 3𝑥2 − 4𝑦2 = 13 tenglamaning natural sonlardagi yechimini toping.
Yechish:
Berilgan tenglamani 4𝑥2 − 4𝑦2 − 12 = 1 + 𝑥2; 4(𝑥2 − y2 − 3) = 1 + 𝑥2
ko‘rinishida yozamiz. Tenglikning chap qismi 4 ga karrali. Har qanday natural sonning kvadratini 4 ga bo‘lganda qoldiqda 0 yoki 1 qolgani uchun 1 + 𝑥2 ifodani 4 ga bo’lganda qoldiq 1 yoki 2 ga teng bo‘ladi, shuning uchun tenglikning o‘ng tomoni 4 ga karrali emas. Demak, berilgan tenglama natural yechimga ega emas.
3‐misol. 𝑥3 − 𝑦3 = 91 tenglamani natural sonlarda yeching.
Yechish. Berilgan tenglamani ko paytuvchilarga ajratamiz
(𝑥 − y)(𝑥2 + 𝑥𝑦 + y2) = 91.
91 sonini 2 ta sonning ko‘paytmasi ko’rinishida yozamiz:
91 = (±1) ⋅ (±91) = (±7) ⋅ (±13) = (±13) ⋅ (±7)
Bu ko paytmalardan foydalanib mumkin bo’lgan barcha tenglamalar sistemalarini tuzamiz. Hisoblashlar suni ko‘rsatadiki, faqat ushbu
𝑥 − 𝑦 = 7
{𝑥2 + 𝑥y + y2 = 13
sistema natural yechimga ega. Bu sistemadan 𝑥 = 5, y = 6 yagona natural yechimni topamiz. Javob: 𝒙 = 𝟓, 𝐲 = 𝟔.
4‐misol. Tenglamaning butun sonlardagi yechimini toping: 𝑥 + 𝑦 = 2𝑥𝑦.
Yechish. Tenglamadan 𝑦 ni topamiz:
𝑥 = 2𝑥𝑦 − 𝑦; 𝑦(2𝑥 − 1) = 𝑥; 𝑦 =
Kasrning butun qismini ajratamiz:
𝑥
2𝑥 − 1
;
2𝑦 =
2𝑥 2𝑥 − 1 + 1
2𝑥 − 1 2𝑥 − 1 2𝑥 − 1 2𝑥 − 1
= = + = 1 +
2𝑥 − 1 1
1
2𝑥 − 1
;
Bundan kelib chiqadiki,
1
2𝑥−1
ifoda butun son bo‘lishi kerak. 1 soni 2𝑥 − 1 ga
bo‘linishi kerak, bundan 2𝑥 − 1 = 1 yoki 2𝑥 − 1 = −1.Tenglamalarni yechib
𝑥 = 0, 𝑦 = 0 yoki 𝑥 = 1, 𝑦 = 1 ekanligini topamiz.
Javob: 𝑥 = 0, 𝑦 = 0 yoki 𝑥 = 1, 𝑦 = 1.
Mustaqil yechish uchun misollar.
I. Tenglamalarni yeching: 1) [𝑥 + 3] = 6
2) [2𝑥 − 3] = 7
3) [3𝑥 − 3] = 𝑥
2
4) [𝑥+3 𝑥
3
2
] =
5) {𝑥 + 3} = 1,6
6) {2𝑥 − 3} = 0,7
7) {2𝑥 − 3} = 𝑥
2
8) {𝑥+3 𝑥
3
2
} =
II. Tenglamalarni butun sonlarda yeching:
1) 2𝑥 + 3𝑦 = 15
2) 3𝑥 + 4𝑦 = 25
3) 2𝑥 − 3𝑦 = 15
4) 𝑥2 − 𝑦2 = 10
5) 𝑥2 − 𝑦2 = 12
6) 𝑥2 − 𝑦2 = 18
7) 𝑥2 − 𝑦2 = 20
8) 𝑥3 − 𝑦3 = 19
9) 𝑥3 − 𝑦3 = 58
10) (𝑥 − 𝑦)3 + (𝑦 − 𝑧)3 + (𝑧 − 𝑥)3 = 9
III. Hisoblang:
1) 𝜏(10)
2) 𝜏(200)
3) 𝜏(300)
4) 𝛿(10)
5) 𝛿(50)
6) 𝛿(100)
7) 𝜑(50)
8) 𝜑(101)
9) 𝜑(𝜏(10))
10) 𝛿(𝜑(100))
11) 𝜑(𝛿(100))