Y e c h i l i s h i . K o ' p h a d d a u mu mi y k o ' p a y t u v c h i y o 'q ,
guruhlash ham mum kin emas. Ammo
-3m
ni
- m - 2m
ko'rinishda
yozsak, k o 'p h a d
m z - m - 2 m + 2
k o ‘rinishga keladi,
endi bu
ko'phadni guruhlab ko'paytuvchilarga ajratish mumkin:
n r - 3m + 2 = n r - m - 2m + 2 - m(m
- 1)
-
2(m - 1) =
= ( m - \ )(m
-
2
).
4.3.
Qisqa ko'paytirish formulalaridan foydalanib ko'paytuvchi
larga ajratish usuli.
M i s o l l a r . Qisqa k o 'p a y tirish form ulalarini k o 'p h a d la rn i
ko'paytuvchilarga ajratishdagi tatbiqini ushbu m isollarda k o 'rib
chiqam iz:
1)
36a2 b4-
2 5= (
6
ab2)2-
5
2
= (
6
ab2-
5)(6
ab2+
5)-
2) 4.v4-
4 x2
+ 1 = (2x2)2- 2 • 2л
-2
+ 1 = (2л2- l)2;
3) 25a6 + 40as + 16
a4
= (5a
1)2
+ 2 • 5a
3
■
4
a2
+
(4a2)2
= (5a
3
+ 4a2)2;
4) 27c3- 0,00 W
/ 6
= (3c)
5
- ( 0 ,0 k
/2)3
= ( 3 c - 0 ,W
2
)(9c
2
+ 0,3
« / 2
+
+
0,0
k / 4);
5) 125 +
8
a
36'2
= 5' + (2
ab4)}
= (5 +
2ab4)(25 -
10
ab4
+
4 a2bs);
6
)
8
- 12c +
6c 2-
с
3
= 2
3
- 3 ■
22c + 3 • 2e
2- c3
= ( 2 - c)3;
7) 6 4 /6+27.-' + 14442 + 1082~ = ( 4 / 2)3+ 3(4rf^. 3r+ 3 - 4
/ 2
■
(32/ +
+ (3r)3= (4
/ 2
+ 3z)3.
B a’zan ko'phadni ko'paytuvchilarga ajratishda bir necha usul-
lardan ketm a-ket foydalanishga to 'g 'ri keladi.
M i s o l l a r . 1) a
3
+ a
2
- 12 = a
3
+ a
2
- 4 -
8
= a
3
-
8
+ a
2
- 4 =
= a3- 2
3
+ a2- 2
2
= (a-2)(a
2
+ 2a + 4) + ( a -
2)(a
+ 2) = ( a - 2)(a
2
+ 2a +
+ 4 + a + 2) = ( a - 2)(a
2
+ 3 a+
6
);
2) 2a
2
- 5a/> + 3/r = 2a
2
-
4ab - a b + 2b2 + b2 - 2a2 - 4ab
+
2b2
+
+ b2- a b
=
2
(/>
2
-
2ab
+ a 2) +
b (b - a )= 2 ( b - a ) 2+ b ( b - a ) = ( b - a )
x
x (
2
(Z>- a) + ft) = (/>-
a ) ( 2 b - 2a+ b) = ( b - a )(3 b -
2
a);
3)m
2- l m + 12= m 2- 3 m - 4 m +
12=
m (m -
3 ) - 4 (w - 3) =
=
( m - 3 ) ( m -
4);
4) ( 4 a - 1)2+ 2 (4 a - 1)+ 1. Agar 4 a - 1 = л belgilashni kiritsak,
berilgan ifoda л
2
+
2
л +
1
ko'rinishga keladi, hosil bo'lgan k o 'p h a d
ni (
1
) form ula yordam ida ko'paytuvchilarga ajratib, oxirgi natijada
x
ning o 'rn iga 4a - 1 ikkihadni qo'yiladi.
( 4 a - 1)2+ 2 (4 a - 1)+
1= x 2+ 2 x +
1 = (л +
1
)
2
= (
4
a _ i +
1)2
=
= (4a
)2
- 16a2.
5 )x
4
+ x
2
+
1
= x 2
+
2
л
2
+
\ — x 2 —
(x
2
+ l)
2
- x
2
= (x
2
+
1
- x) x
x (x
2
+ l + x ) = ( x
2
- x + l)(x
2
+ x +
1
).
48
5-§. Algebraik kasr
T a ’ r i f :
Surat va maxraji algebraik ifodalardan iborat bo 'Igan
kasr algebraik kasr deyiladi.
,
a
a 2+ 5
a x 2 + b x
(
a - 3 ) ( b - 2
)
. ,
.. .
M asalan, — j- , —^— ,
cy
,
----------------- algebraik kasr-
lardir.
Algebraik kasr m axrajining qiymati noldan farqli b o ig a n qiy
m atlarida m a ’noga ega. M asalan,
1
)
“
a +
_ b
h
kasr
a * b
qiym atlarda aniqlangan.
2
) - (
]) k a sr
a
=
0
va
a
=
1
q iy m a tla rd a m a ’n o g a ega
b o im a y d i.
K asrning surat va maxrajini noldan farqli ifodaga k o ‘paytirish
va b o iis h mumkin:
a.
=
6
a.
, bu yerda
с *
0
va
b
0
.
b
c b
5.1.
Algebraik kasrlami qisqartirish.
K asrning
surat va m axra-
jid a ishtirok etuvchi um um iy ko'paytuvchiga surat va m axrajini
b o iis h kasrni
qisqartirish
deyiladi.
m2- n 2 _ (m -n ){m + n ) _ m - n
M i s o l l a r . 1)
-
m(m+n)
~ m
!
m~+mn
4
'
а ^ - 2 а 2Ь _ a 2( a - 2 b ) _ - a 2( 2 h - a ) _
]
^ 2cPb2- a 4b
а*Ь( 2Ь- а)
a 2( 2 b - a ) a b
° b
a 2+b 2+c 2+ 2 a b + 2 b c + 2 a c
(
a + b + c
)2
(
a + b + c ) 2
3) -
a 2- b 2- c 2- 2 b c
a 2- ( b 2+ 2 b c + c 2)
a 2- ( b + c ) 2
(a + b + c ) 2
_ (a + b+c) ( a + b+c )
a + b + c
( a - ( b +c ) ) ( a +b +c ) ~ ( a - b - c ) ( a + b + c ) ~ a - b - c
’
a4+ a 3+ 4a2 +
3 a + 3
a 4+a 3+ a 2 +
3
a 2 + 3 a +
3
4)
я
,
a3- 1
( a - l ) ( a 2+a+ l )
a2(a2+a + l) + 3(a2+ a+ l ) _ (a2 + a + l ) ( a 2 + 3)
a 2 + 3
( a - l ) ( a 2+ o+l )
(a2+a + l )( a - l )
49
5.2.
Algebraik kasrlami qo‘shish.
A lgebraik kasrlam i q o ‘shish
va ayirishda oddiy kasrlarni qo'shish va ayirish am allarini bajarish
kabi avval um um iy m axrajga keltirish kerak.
Buning uchun avval
har bir qo'shiluvchi kasrning maxraji ko'paytuvchiga ajratiladi:
5
3
5
3
5-2
M i s o l l a r .
1
)
2 x - 2 + 4 x - 4
=
2 ( x - \)
+
4 ( x - l ) = 2 - 2 ( x - \) +
3
_
10
3
10 + 3 _
13
+ 4 ( x - l)
~
4 ( x - l) +
4 ( x - \ ) ~ 4 ( x - \ ) ~
4(дг-1) ’
5b-\
b+2 _ b +
1
_
5 b - l
b+2
b +
1
_
5 b - l
2)
зь2- з
+
2b+2
b~ l ~ 3(b2- n
+
2(h+l)
b~ l ~
+
b+2 _ b
+ 1 _
2-(5*-l)
(b + 2 )-3 (b -\ )
(fc+l)-
6
(fc+l)
_
+ 2(6+1)
b-\ ~
2-3(fc-1)(fe+l) + 2(fc+l)-3(fc-l)
( b - l )- 6 ( b + \ ) ~
_ \ o b
- 2
+ з(ь2- ь +
2
ь -
2
)
6 (^ +
2
Л+
1
)
т - г + з ^ + з ь - б - б ^ - т - б
6 ( ^ - 1 )
60?-\)
6 ( ^ - 1 )
6 ( ^ - 1 )
_
- 3 b 2+ b - l 4
6( b2- l )
Aa
cr’+ab _ a - 2
4
a
a ’+ab _ ( a - 2 ) - a ( a +2 )
a
+ 2 +a
2
л
~~ J
*"
2+a
a ( a+2) ~
a(a+ 2)
*"
a^ + 2a
4
a a
a^+ab
_
a ( a 2-
4) +
Aa2- a ^ - a b
a i - A a + Aa2- a i - a b
+ a ( 2 + a )
a ( a +
2
)
~
a( a + 2)
~
a( a+
2
)
~
_
A a 2 - A a - a b _ a ( A a - A - b ) _ A a - b - A
a ( a + 2 )
a ( a +
2
)
a + 2
5.3.
Algebraik kasrlarni ko‘paytirish va bo'lish.
Algebraik kasr
larni k o'paytirish va bo'lish oddiy kasrlarni ko'paytirish va bo'lish
qoidalari bo'yicha bajariladi:
а
с
_
a -с
a
с _ a
d _ a d
Ъ
аГ —
b d '
b
‘
d ~
b
с ~
b
e '
M i s o l l a r .
j.
Am
21k
А- 21- т- к
Ътк
' 9
n
16
d ~
9 1 6
n d ~ And
3
a 2
( a - b ) ( a + b ) - 3 a 2
2)
3 a+3 b 5 b - 5 a
3 ( a + b ) - 5 ( b - a )
2
.2
—15 •
a —b
50
a - b
a - b
a - b
6
b2
2-3
b 2( a - b )
2
^ 9b2
6
b2
9
b 2
a ~ b
3-3
b 2( a - b )
3 ’
АЧ
b2- S b + \ 6
( b -
4)2
( b -
4)2
b2- 9
(/7-4)2 (fo-3)(fe+3)
v — irri— : -------- = i , ^-------------= -----------------------
= b -
3;
fc2- 9
(fc-4)2
(fe+3)(b-4)2
5.4. Algebraik kasrlar ustida birgalikda bajariladigan amallar.
Algebraik kasrlar ustida birgalikda bajariladigan am allar uchun
sonli kasrlar ustida bajariladigan am allarning
tartib va qoidalari
to ‘liq saqlanadi.
M i s o l l a r .
1. Ifodani soddalashtiring:
+
Y e c h i l i s h i . Am allarni bajarilish tartibi b o ‘yicha bajara-
miz. D astlab qavslar ichidagi ifodalarni soddalashtiram iz:
,,
a+b
a - b
_ (
a+b) 2- ( a - b ) 2 __ a 2+2a b +b 2- a 2+ 2 a b - b 2
Aab
' a - b
a+b ~
( a - b ) ( a +b )
~
, , ,
~
7
'
a z - b z
a - b
a-b a+b
_ (
a-b)2+(a+b)2
_
a2-2ab+b2+a2+2ab+b2 _ 2(a2+b2)
2
)
a+b + a-b ~ (a+b)(a-b) ~
,
,
-
’
a*—b
a^-b1
3) Endi b o iis h am alini bajaramiz:
4
ab
1 _ 4
ab
a 2- b 2
_ l a b
a 2- b 2
a 2- b 2 2 ^ a 2+ b 2
j
a 2+b 2 '
4) B o iin m ad a n birni ayiramiz:
2
ab
_
2 a b - a 2- b 2 _
a 2- l a b + b 2 _
( a - b ) 2
a 2+b 2
a 2+ b 2
a 2+b 2
a 2+b
2
T
,
( a - b ) 2
J a v o b : -
a 2+ b 2
2. Ifodani soddalashtiring:
- b2
a x - b _ b x + a
a +b
b - a
a 2- b 2 a 2+ b 2
Y e c h i l i s h i . K o 'p g in a h o lla rd a am allarn in g bajarilish
tartibini yodda tutgan holda ularni birgalikda
ketm a-ket bajarib
borish m aqsadga muvofiqdir:
51
a x - b bx+a
a + b
b - a
‘ a 2- b 2 a 2+b 2 ^
x
2- \ '
* ~ l
(■a x - b ) ( b - a ) - ( b x + a ) ( b + a
)
(
b + a ) ( b - a
)
a 2- b 2
x —
1
a b x - a 2x - b 2+ a b - b 2x - a b x - a b - a 2
a 2 + b
2
b 2- a 2
p + b
2
W l )
-д
^ a 2+b2
j - ^ a 2+62 j
a2- b 2
(a2+b2 )
^+1)
a 2- h
2
[a2+b2 y +
1)
|
(a2+b2 )\x+\)
J a v o b :
1
.
3. Ifodani soddalashtiring va uning
a
= 0,5 dagi qiym atini hi-
„2 4
soblang
1
+
3 x + j r
x + 3
a x - 2 a 2
x 2+ x - 2 a x - 2 a
Y e c h i l i s h i : 1) Ifodani soddalashtiram iz:
(
x
2
a x - 2 a 2
x 2+ x - 2 a x - 2 a
Здг+л:2 | _
x
2
+ x + 3 \ ~ a ( x - 2 a )
л(д:+1)-2а(дг+1) x
(*+3)(*+l)
x+3+x(x+3)
x
2
x+3
~ a (x -2 a )
(
x+ \)(x-2 a )
x+3
~
a(x-2a )
2
_
x -2 a _
1
x -2 a ~ a (x-2 a ) ~ a
•
2)
Ifodaning
a
= 0,5 dagi qiym atini hisoblaymiz:
1
1
0 a =
0,5
J a v o b :
2
.
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