2. Mantiqiy masalalarni yechishning asosiy usullari

Teorema. G (a)  H(a) tengsizlik to‘g’ri , jumladan, H(a) = G(a) tenglik faqat va faqat a1=a2 =…= an bo‘lganda bajariladi.
Isboti. Koshi tengsizligidan foydalanib (1-masalaga qarang) foydalanib
(H(a)) -1= a1
 n a1a1...a1  (G(a))1 tenglikka ega bo‘lamiz.
1 2 n
 a1  ...  a1
2 n
1
n
Jumladan, H(a) = G(a) tenglik faqat a1=a2 =…= an da bajariladi.
1-misol. Agar a, b, c  0 bo‘lsa,
3
1 a 1 b 1 c 3
 a  b  c
tengsizlikni
isbotlang.
Yechilishi: 9  a  b  c 1  1  1  tengsizlikni isbotlaymiz:
 a c 
b
 
 a  b  c 1  1  1   9
a  b  c  33 abc ,
3
   3 .  a b c  3 abc
3 abc
abc  9.

 1 1 1 1
 a b c
 

2-misol. Agar a, b, c  0 , ab2c3  1 bo‘lsa, 1  2  3  6 ni isbotlang.
a b c
Yechilishi: 1  2  3  1  1  1  1  1  1  6
a b c a b b c c c
6 ab2c3
1
 6 .
Mustaqil yechish uchun misollar
1. x, y, z  0 bo‘lsa, u holda quyidagi tengsizlikni isbotlang:
 x  y  z 2 9 xyz  x  y  z 
x2 y2 z2
1 1 1 1 28 3
    .
2. Agar x1, x2 , ..., xn  0 va x1  x2 ...  xn  1 bo‘lsa, u holda quyidagi tengsizlikni isbotlang:
1 x1  x2 ...  xn  1 x1 ...  xn1  xn
x1 x2
x1  x2  ...  xn
 ... 
 1
xn
.
3. x, y, z  0 bo‘lsa, u holda quyidagi tengsizlikni isbotlang:
2 2 2
x  xy  y  yz  z  xz  0.
x  y
y  z
x  z
4. Agar a, b, c  0 bo‘lsa, u holda quyidagi tengsizlikni isbotlang:
   2 .
a b
c
b  c a  c a  b
5. Agar a, b, c  0 bo‘lsa, u holda quyidagi tengsizlikni isbotlang:
ap2  bp2  cp2  apbc  bpac  c pab .
8.Funksional tenglamalarni yechishning asosiy usullari.
Funksional tenglamalar haqida tushuncha.
Noma’lum funksiyaga nisbatan qaralayotgan tenglama funksional tenglama deyiladu. Masalan bir o‘zgaruvchili funksiyaning juftlik, toqlik, qo‘zg’almas nuqtaga ega bo‘lishi xossalarini ifodalovchi funksional tenglamalari mos ravishda
f (x)  f (x), f (x)  f (x), f (f (x))  x
tenglamalardan iborat. Shu kabi
f (x)  f  x , f (x)  cos 1  f  x , f (0)  1 ,
 2   2 
2
f  uzluksiz funksiya.
   
tenglamalar bir o‘zgaruvchili funksional tenglamalarga misol bo‘ladi. o‘zgaruvchili
f (x  y)  f (x)  f ( y); f (x  y)  f (x) f ( y);
f (xy)  f (x)  f ( y); f (xy)  f (x) f ( y);
funksional tenglamalar Koshi tenglamalari deyiladi. Shu bilan birga ikki o‘zgaruvchili funksional tenglamalarga
Ikki
f (x  y)  f (x)  f ( y) 
2
Iyensen tenglamasini va
f (x  y)  f (x  y)  2 f (x) f ( y)  Dalamber tenglamasini
misol keltirishimiz mumkin. Shuningdek, f (x)  sin x, g(x)  cos x trigonometrik funksiyalarni qo‘shish fomiulalari asosida tuzilgan ikki noma’lumli ikki o‘zgaruvchili
f (x  y)  f (x)g( y)  g(x) f ( y);
f (x  y)  f (x)g( y)  g(x) f ( y);
g(x  y)  g(x)g( y)  f (x) f ( y);
g(x  y)  g(x)g( y)  f (x) f ( y);
funksional tenglamalarni yozishimiz mumkin. Yana misol sifatida trigonometrik funksiyalarni qo‘shish formulalari asosida
f (x)  tgx
f (x  y)  f ( x)  f ( y) ; f (x  y)  f ( x)  f ( y)
;
1 f (x)  f ( y) 1 f (x)  f ( y)
funksional tenglamalarni yozishimiz mumkin.
Odatda funksional tenglamalar ko‘plab yechimlarga ega bo‘ladi va bu tenglamalarni bevosita yechish qiyinchilik tug’diradi. Ammo izlanayotgan funksiyaning ba’zi xarakteristik xossalari: uzluksizligi, davriyligi, chegaralanganligi kabi xossalari berilsa funksional tenglamani yechish osonlashadi. Ko‘plab matematik turnir va matematik olimpiadalarda funksional
tenglamalarni yechishga oid misollar berilmoqda. Aytilganlarni e’tiborga olgan holda ba’zi klassik funksional tenglamalarni yechilishini ko‘rib o‘tamiz.
1-misol. Agar
f  3x 1   x 1 bo‘lsa,
 x  2  x 1
 
f (x) funksiyani toping.
Yechilishi. Berilgan tenglamani yechish uchun 3x 1  t almashtirish kiritamiz. Bu
x  2
tenglikdan x ni topamiz:
3x 1  t(x  2)
3x  tx  2t 1 (3  t)x  2t 1
x  2t 1 (t  3).
3  t
Berilgan tenglamaning o‘ng tomonidagi x lar o‘rniga t orqali ifodasini qo‘yamiz va ifodani soddalashtiramiz:
2t 1 1
x 1  3  t  2t 1  3  t  t  4
.
2t 1 1
x 1
2t 1 3  t 3t  2
3  t
Berilgan tenglama f (t)  t  4 ko‘rinishga keladi. Erkli o‘zgaruvchi bo‘lgan t ni x
3t  2
ga almashtirsak qidirilayotgan funksiya topiladi.
Javob:__f_(_x_)__x__4_(_x__3)._3_x__2_2-misol.'>Javob:
f (x)  x  4 (x  3).
3x  2
2-misol. Agar f (x)  2 f  1   x bo‘lsa f (x) funksiyani toping.
 x 
 
Yechilishi. Bu tenglamani yechish uchun x lar o‘rniga 1 ni qo‘yamiz, ya’ni x  1
x x
almashtirish bajaramiz. Berilgan tenglama
f  1   2 f (x)  1 ko‘rinishga keladi.
 x 
x
 
Endi ikkita tenglamadan sistema tuzib olamiz:
  1 
f (x)  2 f  x
(1)
 f  1   2 f (x)  1
(2)
 x 
 x 
x

  


  
(2) tenglamani 2 ga ko‘paytirib birinchisidan ayiramiz va f (x) ni topamiz:
  1 
f (x)  2 f  x
2 f  1   4 f (x)  2
 x 
 x 
x

  


 

3 f (x)  x  2
x
x2  2
f (x) 
3x
Javob:
x2  2
f (x) 
3x
.
3-misol. Agar (x 1) f (x)  f  1  
1
x 1
 x 
 
bo‘lsa f (x) funksiyani toping.
Yechish: Bu tenglamani yechish uchun ham x  1 almashtirish bajarilsa u
x
ga ko‘paytiramiz( x  0, x  1).
 1 1 f  1   f (x) 
1 x
x
 x   x 

  
ko‘rinishni oladi,
1  x
x
2
f  1   x
f (x)  
1 x
 1 x 
x  .
 x 
   
Hosil qilingan tenglamani berilgan tenglama bilan birgalikda yechamiz.
2
  1  1
(x 1) f (x)  f 
x 1
 f  1   x f (x)   x 
1 x
 1 x 
 x 
 x 

 



  


2
1
 x 1
 f (x) 
1 x 
x 1  1 x 
x
  x 





1
f (x) 
1 x
.
Javob: f (x) 
1
, x  0.
1 x
4-misol. Agar f (x)  xf 
  2 bo‘lsa f (x) funksiyani toping.
 2x 1 
x
 
Yechish:
2x 1
x
 t belgilash kiritamiz. Bundan x ni topamiz
x  t(2x 1)
x(1 2t)  t
,t  1 .
2t 1 2
t
x 
va berilgan tenglama
f (t)  2 ko‘rinishga keladi. t  x almashtirish
 2t 1  2t 1
f  t   t


bajaramiz:
f (x)  2 . Hosil qilingan tenglamani x ga ko‘paytiramiz
 2x 1  2x 1
x
x
f 
 
 
va berilgan tenglamani ayirib, f (x) ni topamiz:
x2
1 f (x)  2x  2

 2x 1



(x 1)2
f (x)  2(x 1)
2x 1
f (x)  4x  2 (x  1)
x 1
ni hosil qilamiz.
x  1da berilgan tenglamadan f 1  1
Javob: f (x)   x 1
 4x  2 , x  1, x  1 ,
2
1, x  1.
Mustaqil yechish uchun misollar.


• Agar
• Agar 3)

f  x  2  x  2 bo‘lsa,
f (x) funksiyani toping.
f 2x 1  3x  4 bo‘lsa,
f (x) funksiyani toping.
Agar f x2  2x 1  x2  2x  4 bo‘lsa, f (x) funksiyani toping.
4) Agar
f  x  2   x 1  x bo‘lsa,
3 2
 


f (x) funksiyani toping.
5) Agar
1
1
 x  2  x 1
f 
 


bo‘lsa, f (x) funksiyani toping.
6) Agar
f  3x 1   x 1 bo‘lsa,
 x  2  x 1


f (x) funksiyani toping.
7)
Agar f  x 1   x  2 1 bo‘lsa, f (x) funksiyani toping.
 x  2  x 1
 
8)
Agar 2 f (x)  f  1   x bo‘lsa f (x) funksiyani toping.
 x 
 
9) Agar 2 f (x)  3 f  1   x 1 bo‘lsa
 x 
 
f (x) funksiyani toping.
10) Agar 2 f (x  1)  3 f 
1
  x  1 bo‘lsa
 x  1 


f (x) funksiyani toping.
11) Agar (x 1) f (x)  f  1  
1
x 1
 x 
 
bo‘lsa f (x) funksiyani toping.
12) Agar xf (x)  f 
  2 bo‘lsa f (x) funksiyani toping.
 2x 1 
x


13) Agar
f (x)  2xf 
  3 bo‘lsa f (x) funksiyani toping.
 2x 1 
x
 
14) Agar f (x)  f  2x 1   2 bo‘lsa f (x) funksiyani toping.
 x  2 
 
15) Agar xf ( 3x 1)  2 f  x  x bo‘lsa f (x) funksiyani toping.
2x  3