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(Viet teoremasi);

1 2 _a 1 2 _a
2. ax²  bx  c  a(x  x₁)(x  x₂ ) ;

2 2 ²
b²  2ac

^{3. x1  x2 }  ^{x1  x2} 
3
 2x₁x₂  _a2 ;
_ b _³ 3bc

^{4. x3  x3 }  ^{x  x}  ^{ 3x x (x  x )    ;}
1 2 1 2 1 2 1 2  

a

_a2
 
1 1 b²  2ac 1 1 b³  3abc
5. _x2  _x2  _c2 ; _x3  _x3  _c3 ;


1 2 1 2
 ^b ²
b²  4ac

6. to‘la kvadratga ajratish:
ax²  bx  c  a _ x 

2a ^{ } 4a ^.

Agar
x₁,
^x2,

Kub tenglama

x³  ax²  bx  c  0 .
x₃ lar bu tenglamaning ildizlari bo‘lsa, u holda

1. x³  ax²  bx  c  (x  x₁)(x  x₂ )(x  x₃ ) .
 ^{x1  x2  x3  a,}
2. Viet teoremasi: ^ x x  x x  x x  b,
^ 1 2 1 3 2 3
^ x x x  c.
 1 2 3

Ikki noma’lumli chiziqli tenglamalar sistemasi

Umumiy ko‘rinishi:
_a₁₁x  a₁₂ y  b₁,
^a x  a y  b .

Geometrik talqini

1. 
   Agar
   

^a11 ^a21


_ ^a12
^a22



_ b₁ b₂
21 22 2
bo‘lsa,

sistema yechimga ega emas.

2. 
   Agar
   

^a11 ^a21
_ ^a12
^a22
bo‘lsa,

sistema yagona yechimga ega.

3. 
   Agar
   

^a11 ^a21
_ ^a12
^a22
_ b₁ b₂
bo‘lsa,

sistema cheksiz ko‘p yechimga ega.

Arifmetik progressiya

a_n1  a_n  d ,
n  0, 1, 2,... .

bu yerda
a₁ – birinchi hadi, d –ayirmasi.

1. 
   ayirmasini topish:
   

d  a_n1

• 
  a_n
  

 ^{an  am} , n  m ;
n  m

2. 
   n - hadini topish:
   

^{an  a1 } ^{n 1}^{d ;}

3. 
   o‘rta hadini topish:
   

_{a } ^ank ^{ a}nk n ₂
, k  n ;

4. 
   dastlabki n ta hadining yig‘indisini topish:
   

_{S } a₁  a_{n  n } ^{2a1 } ^{n 1}^d _{ n ;}
ⁿ 2 2

5. 
   n - dan k - gacha bo‘lgan hadlar yig‘indisini topish:
   

^{Sk  a  dn(k 1)} ^{k  n} ^;

6. 
   a_n  S_n  S_n1 .
   

n n
Geometrik progressiya b_n1  b_nq, n  1, 2, 3,... .

bu yerda,
b₁ – birinchi hadi, q – maxraji.

1. 
   maxrajini topish:
   
2. 
   n - hadini topish:
   

_{q } ^bn1 _;
b_n
b_n  b₁  q^n1, b_n  b_nm  q^m ;

3. 
   o‘rta hadini topish:
   

b_n  ;

4. 
   dastlabki n ta hadi yig‘indisini topish:
   

b q  b
b_{1 }qⁿ 1_

1. 
   b_n  S_n  S_n1 ;
   

S_n 
n 1 _
q 1
q 1 ^;

2. 
   Cheksiz kamayuvchi geometrik progressiya hadlari yig‘indisi:
   


S ^b1 ,
1  q
q  1.

Tengsizliklarning xossalari

1. 
   Agar
   

f (x)  g(x) bo‘lsa,
c  0 da cf (x)  cg(x) ,
c  0 da cf (x)  cg(x)
bo‘ladi.

2. 
   Agar bo‘ladi.
   

f ²ⁿ (x)  g ²ⁿ (x) _ f ²ⁿ(x)  g²ⁿ (x)_
bo‘lsa, u holda
f (x)  g(x) _ f (x)  g(x) _

3. 
   Agar o‘rinli.
   

f (x)  c _ f (x)  c _
bo‘lsa, u holda c  f (x)  c
 ^{f (x)  c ёки
f (x)  c}
lar

_1. f (x) _{ 0}
g(x)

Kasr ratsional tenglama va tengsizliklar

_ f (x)  0,


^{ }g(x)  0.

_2. f (x) _ h(x)
_{ } f (x)v(x)  g(x)h(x)  0,

g(x) v(x)
^g(x)v(x)  0.


3. ^{f (x)}  0 
g(x)


f (x)
_ f (x)g(x)  0,


^g(x)  0.
^ ^{f (x)  g(x)h(x)} ^{g(x)  0,}

4.  h(x) g(x)
 _
g(x)  0.

Agar a  b  c  d

bo‘lsa, u holda

Oraliqlar usuli

a) (x  a)(x  b)(x  c)(x  d)  0
bo‘ladi.
tengsizlikning yechimi (;a) (b;c) (d;)

b) (x  a)(x  b)(x  c)(x  d )  0
Bu usul oraliqlar usuli deyiladi.
tengsizlikning yechimi (a;b)  (c;d )
bo‘ladi.

Kvadrat tengsizlik

ax²  bx  c  0 _ax²  bx  c  0_

1) a  0, D  0 bo‘lsa,	^x^^^;^x^1 ^^x^2;^	x x1; x2 ;
2) a  0, D  0 bo‘lsa,	^x^^^;^	 ^x^^x^1 ^;
3) a  0, D  0 bo‘lsa,	^x^^^;^	 ^x^ ^;
4) a  0, D  0 bo‘lsa,	^x^^x^1;^x^2	; x1  x2 ; ;

5) a  0, D  0
bo‘lsa,
x  x_1
x ;_ ;

6) a  0, D  0
bo‘lsa, x 
_x ;_ .

Irratsianal tenglama va tengsizliklar

1. ^2k f (x)  0
 f (x)  0 .
_g(x)  0,

2. 
   2k
   

^{f (x)  g}  ^x
 _
 f (x)  g ^2k

(x).

2. 
   2k
   

^{f (x)  2k g}  ^x 
_{ } f (x)  0,

^ f (x)  g(x).


^{4. 2k1 f (x)  g}  ^x
 f (x)  g ^2k1(x) .
_ f (x)  0, g(x)  0,

_5. 2k
^{f (x)  g}  ^x
 _
 f (x)  g ^2k

(x).

^{6. 2k1 f (x)  g}  ^x
 f (x)  g ^2k1(x) .
_g(x)  0,
_g(x)  0,

_7. 2k
^{f (x)  g}  ^x
 _
 f (x)  g ^2k

(x).

ёки
^ f (x)  0.


^{8. 2k1 f (x)  g}  ^x
 f (x)  g ^2k1(x) .