Qoldiqli bo‘lish
a p q r (0 r p) ,
bu yerda a –bo‘linuvchi, p –bo‘luvchi, q –bo‘linma, r –qoldiq.
O‘lchov birliklari
1. 1 km=1000m. 7. 1m3=1000dm3=1000000sm3.
2. 1 m=10 dm=100sm. 8. 1dm3=1000sm3.
3. 1 sm=10 mm. 9. 1l=1dm3=1000sm3.
4. 1 km2=1000000m2. 10. 1t=10s=1000kg.
5. 1 m2=100dm2=10000sm2. 11. 1kg=1000g.
6. 1 ga=100ar=10000m2. 12. 1g=1000mg.
Oddiy kasrlar ustida amallar
a c a c ; a c
ad bc ; a c
a c ;
a : c
a d
a d .
b b b
b d bd
b d b d
b d b c b c
Agar
a : b c : d
Proporsiya
bo‘lsa, quyidagilar o‘rinli:
ad bc ; 2) ma nb mc nd ;
pa qb pc qd
3) a b b ; 4) a c
a ;
c d d b d b
Davriy o‘nli kasrlarni oddiy kasrga aylantirish
a , a a a a
(b b b b
) a0a1a2a3....akb1b2b3....bn a0a1a2a3 ak
0 1 2 3
k 1 2 3 n
99–9...9–9900–0...0–00
n k
Sonli tengsizliklar
a b
a b 0 , a b
b a ;
a b,
a b,
a b,
b c с 0
с 0
a c , a b
ac bc, a b ;
c c
ac bc, a b ;
c c
a с b c ;
6) a b 0,
c 0
a b , c c .
c c a b
Agar
x1, x2 ,..., xn
O‘rta qiymatlar
musbat sonlar bo‘lsa, ularning:
o‘rta arifmetigi:
A x1 x2 ... xn ;
n
o‘rta garmonigi: H
1
x1
n ;
x
x
1 ... 1
2 n
x 2 x 2 ... x 2
o‘rta kvadratigi: K 1 2 n ;
n
Koshi teoremasi: H G A K .
Protsentlar
1) a sonining P protsentini topish:
x a P ;
100
P protsenti a ga teng sonni topish:
x a 100 ;
P
a soni b sonining
a 100%
b
ini tashkil etadi;
a soni a + b yig‘indining
a a b
100%
ini tashkil etadi.
Daraja va uning xossalari
1) an a–a–a – – a ; 2)
n
a0 1,
a1 a ;
3) a p 1
a p
; 4)
a p aq a pq ;
a p : aq apq ; 6) a p q a pq ;
7) a bp
a p
ap bp ; 8)
b
a p
bp .
Bu yerda
n N ,
a 0,
b 0 ;
q, p R .
Haqiqiy sonning moduli
a
a, agar
a 0 bo'lsa;
1) ab
a b ;
a, agar
2)
a 0 bo'lsa.
;
3) a b
a b ; 4)
a b
a b ; 5)
a 2 a2 .
Arifmetik ildiz va uning xossalari
a 0,
n a
0,
a ,
n 2k, k N ,
n
n a
a.
a,
n 2k 1,
k N.
1) ; 2) ;
n
3) am ; 4) mp anp
5) an a p ; 6)
7) mnp anpbpc ; 8)
m an ;
mn a ;
n1 a ;
9)
A m ; m .
2
Qisqa ko‘paytirish formulalari
1) a b2 a2 2ab b2 ; 3) a b3 a3 3a2b 3ab2 b3 ;
2) a2 b2 a b a b ; 4)
5) a4 b4 a b a b a2 b2 ; 6)
a3 b3 a b a2 ∓ ab b2 ;
(a b c)2 a2 b2 c2 2(ab ac bc) .
Kombinatorika elementlari
m ta elementdan n tadan barcha o‘rinlashtirishlar soni:
m
A
n m(m 1)(m 2)...(m n 1)
m! .
(m n)!
n ta elementdan barcha o‘rin almashtirishlari soni:
Pn n! 1 2 3 4 ... n .
m ta elementdan n tadan barcha gruppalashlar soni:
A
m
n
C
n m m(m 1)(m 2)...(m n 1)
Pn 1 2 3 n
m! .
(m n)! n!
Chiziqli tenglama
ax b 0 .
1. a 0,
2. a 0,
b R b 0
bo‘lsa, yagona yechimga ega: bo‘lsa, yechimi yo‘q;
x b ;
a
3. a 0,
b 0
bo‘lsa, yechimi cheksiz ko‘p: x R .
Kvadrat tenglama
ax2 bx c 0 ( a 0 ).
1. D b2 4ac 0
bo‘lsa, 2 ta har xil haqiqiy yechimi bor:
b
x1,2 2a .
2. D b2 4ac 0
bo‘lsa, 1 ta ikki karrali haqiqiy yechimi bor:
x b .
3. D b2 4ac 0
1,2 2a
bo‘lsa, haqiqiy yechimi yo‘q.
Yechimlarining xossalari:
Agar
x1 va
x2 sonlar
ax2 bx c 0
tenglamaning ildizlari bo‘lsa, u holda:
x x b , x x c
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