(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,
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
Umumiy ko‘rinishi:
a11x a12 y b1,
a x a y b .
Geometrik talqini
Agar
a11 a21
a12
a22
b1 b2
21 22 2
bo‘lsa,
Agar
a11 a21
a12
a22
bo‘lsa,
sistema yagona yechimga ega.
Agar
a11 a21
a12
a22
b1 b2
bo‘lsa,
sistema cheksiz ko‘p yechimga ega.
an1 an d ,
n 0, 1, 2,... .
bu yerda
a1 – birinchi hadi, d –ayirmasi.
ayirmasini topish:
d an1
an am , n m ;
n m
n - hadini topish:
an a1 n 1 d ;
o‘rta hadini topish:
a an k an k n 2
, k n ;
dastlabki n ta hadining yig‘indisini topish:
S a1 an n 2a1 n 1 d n ;
n 2 2
n - dan k - gacha bo‘lgan hadlar yig‘indisini topish:
Sk a dn(k 1) k n ;
an Sn Sn1 .
n n
Geometrik progressiya bn1 bnq, n 1, 2, 3,... .
bu yerda,
b1 – birinchi hadi, q – maxraji.
maxrajini topish:
n - hadini topish:
q bn1 ;
bn
bn b1 qn1, bn bnm qm ;
o‘rta hadini topish:
bn ;
dastlabki n ta hadi yig‘indisini topish:
b q b
b1 qn 1
bn Sn Sn1 ;
Sn
n 1
q 1
q 1 ;
Cheksiz kamayuvchi geometrik progressiya hadlari yig‘indisi:
S b1 ,
1 q
q 1.
Agar
f ( x) g( x) bo‘lsa,
c 0 da cf ( x) cg( x) ,
c 0 da cf ( x) cg( x)
bo‘ladi.
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)
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)
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,
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x ; x1 x2;
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x x1; x2 ;
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2) a 0, D 0 bo‘lsa,
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x ;
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x x1 ;
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3) a 0, D 0 bo‘lsa,
|
x ;
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x ;
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4) a 0, D 0 bo‘lsa,
|
x x1; x2
|
; x1 x2 ; ;
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5) a 0, D 0
bo‘lsa,
x x1
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,
2k
f (x) g x
f ( x) g 2k
(x).
2k
f (x) 2k g x
f ( x) 0,
f (x) g(x).
4. 2k1 f (x) g x
f (x) g 2k1(x) .
f (x) 0, g(x) 0,
5. 2 k
f (x) g x
f ( x) g 2k
(x).
6. 2k1 f (x) g x
f (x) g 2k1(x) .
g(x) 0,
g(x) 0,
7. 2 k
f (x) g x
f ( x) g 2k
(x).
ёки
f ( x) 0.
8. 2k1 f (x) g x
f (x) g 2k1(x) .
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