Question
12. \( \frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}=3 \) tenglamani yeching. A) \( x=a b c \) B) \( x=a+b+c \) C) \( \emptyset \) D) \( x=1 \) 13. Arifmetik progressiyada \( a_{m+n}=x \) va \( a_{m-n}= \) \( y \) bo'lsa, \( a_{m} \) ni toping. A) \( x \cdot y \) B) \( \frac{x-y}{2} \) C) \( \frac{x+y}{2} \) D) \( x+y \) 14. \( b \) ning qanday qiymátlarida \( y=b^{2}(x+1)+ \) \( 4 b(x-1)+5 x+9 \) chiziqli funksiya o'suvchi bo'ladi? A) \( b \in(-5 ;-1) \) C) \( b \in(-4 ;-1) \) B) \( b \in(1 ; 5) \) D) \( b \in R \)
Ask by Salinas Hobbs. in Uzbekistan
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
**12. Tenglama:**
\[
\frac{x - a - b}{c} + \frac{x - b - c}{a} + \frac{x - c - a}{b} = 3
\]
**To'g'ri javob:**
**B) \( x = a + b + c \)**
---
**13. Masala:**
Arifmetik progressiyada:
\[
a_{m+n} = x \quad \text{va} \quad a_{m-n} = y
\]
bo'lsa, \( a_m \) ni toping.
**To'g'ri javob:**
**C) \( \frac{x + y}{2} \)**
---
**14. Masala:**
Funksiya:
\[
y = b^{2}(x + 1) + 4b(x - 1) + 5x + 9
\]
qachon o'suvchi bo'ladi?
**To'g'ri javob:**
**D) \( b \in \mathbb{R} \)**
Solution
**12. Tenglama:**
\[
\frac{x - a - b}{c} + \frac{x - b - c}{a} + \frac{x - c - a}{b} = 3
\]
**Yeșilish:**
Har bir bo'linmani alohida ko'rib chiqamiz:
\[
\frac{x - a - b}{c} = \frac{x}{c} - \frac{a + b}{c}
\]
\[
\frac{x - b - c}{a} = \frac{x}{a} - \frac{b + c}{a}
\]
\[
\frac{x - c - a}{b} = \frac{x}{b} - \frac{c + a}{b}
\]
Ularni qo'shganda:
\[
\frac{x}{c} + \frac{x}{a} + \frac{x}{b} - \left( \frac{a + b}{c} + \frac{b + c}{a} + \frac{c + a}{b} \right) = 3
\]
Bu yerda ko'rinib turibdiki:
\[
x \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) - \left( \frac{a + b}{c} + \frac{b + c}{a} + \frac{c + a}{b} \right) = 3
\]
Agar \( x = a + b + c \) deb qo'ysak:
\[
(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) - \left( \frac{a + b}{c} + \frac{b + c}{a} + \frac{c + a}{b} \right) = 3
\]
Bu tenglama to'g'ri keladi.
**To'g'ri javob:**
**B) \( x = a + b + c \)**
---
**13. Masala:**
Arifmetik progressiyada:
\[
a_{m+n} = x \quad \text{va} \quad a_{m-n} = y
\]
bo'lsa, \( a_m \) ni toping.
**Yeșilish:**
Arifmetik progressiyada istalgan ikkita tenglama bo'yicha:
\[
a_{m+n} + a_{m-n} = 2a_m
\]
Demak:
\[
x + y = 2a_m \implies a_m = \frac{x + y}{2}
\]
**To'g'ri javob:**
**C) \( \frac{x + y}{2} \)**
---
**14. Masala:**
Funksiya:
\[
y = b^{2}(x + 1) + 4b(x - 1) + 5x + 9
\]
qachon o'suvchi bo'ladi?
**Yeșilish:**
Funksiyani soddalashtiramiz:
\[
y = b^2 x + b^2 + 4b x - 4b + 5x + 9 = (b^2 + 4b + 5)x + (b^2 - 4b + 9)
\]
Bu chiziqli funksiya bo'lib, o'suvchi bo'lishi uchun koeffitsient \( x \) ga nisbatan musbat bo'lishi kerak:
\[
b^2 + 4b + 5 > 0
\]
Diskriminant:
\[
D = 16 - 20 = -4 < 0
\]
Demak, kvadrat har doim musbat bo'ladi, ya'ni har qanday \( b \) uchun \( y \) o'suvchi.
**To'g'ri javob:**
**D) \( b \in \mathbb{R} \)**
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Simplify this solution Beyond the Answer
Bu tenglamani yechishdan avval, har bir kesmani \( x \) ga nisbatan tartibga keltirish va oddiy emas, balki tenglama aylanmasi uchun ko'paytirilishi kerak. Shunday qilib, juda ko'p oddiy manipulyatsiyalar orqali siz oxir-oqibat \( x=a+b+c \) natijasiga kelasiz. Arifmetik progressiyada, elementlar orasidagi farq doimiy bo’ladi. Shunday ekan, \( a_{m} \) ni topish uchun, \( a_{m+n} = x \) va \( a_{m-n} = y \) ni hisobga olib, arifmetik progressiyaning xususiyatlarini qo’llagan holda, natijaga kelishingiz mumkin: \( a_{m} = \frac{x+y}{2} \).
