2. Mantiqiy masalalarni yechishning asosiy usullari

Javob: tenglama yechimga ega emas.
7-misol. Tenglamani yeching. 1 x  x2  2x2  5x  2
Yechish.
a1  x , a2  1, a3  1, b1  x, b2  x, b3  1
ko‘rinishda
belgilab olamiz. U holda

2
 2 
2 2 2 
 1  x  x2   x  1  1 x  x  1  2x2  5x  1
2 2
a1b1  a2b2  anbn  a1  a2

• an b1  b2  bn 

a1 a2 a3
tenglik faqat   da bajarilishini hisobga olsak,
b1 b2 b3
ekanini topish mumkin.
Javob: x  1
Tenglamaning butun yechimlarini topish.
x  1
x x
 1  x  1
1
Tenglamaning butun yechimlarini topish uchun quyidagi teoremadan foydalanamiz:
𝑥 = 𝑝 qisqarmas kasr butun koeffitsiyentli 𝑎 𝑥𝑛 + ⋯ + 𝑎 𝑥 + 𝑎 = 0, 𝑎 ≠ 0 ,
𝑞
𝑛 1 0 𝑛
tenglamaning ildizi bo‘lishi uchun p soni 𝑎0 ozod hading, q esa 𝑎𝑛 bosh koeffitsiyentning bo‘luvchisi bo‘lishi zarur.
𝑝
Bundan butun yechimlarni ko‘rinishdagi butun sonlardan qidirish kerakligi kelib
𝑞
chiqadi.
8-misol. 𝑥3 − 3𝑥 + 2 = 0 tenglamaning butun ildizlarini toping. Yechish: Ozod had -2 ga , bosh koeffitsiyent 1 ga teng. p soni -2 ning bo‘luvcilari: ±1, ±2. q soni 1 ning bo‘luvchilari: ±1. 𝑝 = ±1, ±2, 𝑞 = ±1 .
𝑥 = 𝑝 = ±1, ±2 . 𝑥 ning qiymatlarini tenglamaga qo‘yib tekshirib ko‘ramiz.
𝑞
𝑥 = −1 da (−1)3 − 3 ∙ (−1) + 2 ≠ 0 ,
𝑥 = 1 da 13 − 3 ∙ 1 + 2 = 0 ,
𝑥 = −2 da (−2)3 − 3 ∙ (−2) + 2 ≠ 0 ,
𝑥 = 2 da 23 − 3 ∙ 2 + 2 ≠ 0 .
Javob: 𝑥 = 1.
Mustaqil yechish uchun misollar.

• Tenglamani yeching. √𝑥 − 4 + √4 − 𝑥 = 0.
• Tenglamani yeching. √5 − 𝑥 + √𝑥 − 10 = 2.

3) Tenglamani yeching. 2√𝑥 − 4 + 3√4 − 𝑥 = 𝑥3 − 64.

• Tenglamani yeching. √𝑥 − 4 + √𝑥 − 1 = 1,5
• x  5  lg25  x2 tenglamani yeching.
• Tenglamaning butun yechimlarini toping √𝑥 − 4 + √10 − 𝑥 = 2√3.
• Tenglamaning butun yechimlarini toping √𝑥 − 2 + 4√10 − 𝑥 = 2 + √2
• Tenglamani yeching. sin 2x 1  x2  2x  3 .
• x  3x  81 tenglamani yeching.
• (x 1)  4x1  4 tenglamani yeching.
• Tenglamani yeching.

a, b  R, ab  x  y 2  x2  a2  1 y2  b2  1.

• Tenglamani yeching. x  2 y  z 2  9x2  y2  z2 .

• Tenglamani yeching. 3 x  y  z  
• Tenglamalar sistemasini yeching.



y  z  x .
z  2  x  y  z ,
5 x  y   4 z  1,   0,  .
x sin   y cos 
2 


 


15) Tenglamani yeching.
  9, 0  a  x  b
 a  b  b  x  x  a  x a
b
x a
b  a  b b  x x  a 
 



