|
b) x t
у - Х У
1
y a _ xc
+
y a '
а
с
а с
с a
ca
2
. a)
4
1
4 -9
1-7
36
7
3 6 - 7
29
7
9 “ 7-9
9-7
63
63 "
63
~ 6 3 ’
a
c
a d
c d _ ad - c b
) b
d ~ ~ b ~ d + 7 ~ b ~
ad
’
5 3 _ 5 3
5
а с
ac
3 a > 6 4 = (T~4 ~ 8 ’
b)
b 'd = b d ;
л
\
Z - 2 8 - Z
36 _ 4 _ ,
a
с
a d
ad
4* a) 9* 36
9 28 ~ 4
’
b) V
d ~ ~ b '~ c ~ T c '
Yuqorida o'xshash misollarni ko‘ rsatgandan so‘ ng o'qituvchi yana
bir ayniy almashtirishning mazmunini quyidagicha tushuntirishi lozim.
Har qanday ayniy almashtirishning maqsadi misol yoki masalani yechish
uchun berilgan matematik ifodani eng sodda yoki qulay holatga keltirib
hisoblashdan iboratdir.
_
. ,
a2 - 25
a
a + 5
l-misol.
-
-----
5
— ------- =------ ifodani soddalashtiring.
a + 3
a2 + 5a
a2 + 3a
Y e c h i s h :
?2 - 25
a
= a2 - 52
a
_ ( a - 5 ) ( a + 5)
a
a - 5,
a + 3
a2 + 5a
a + 3
a(a + 5) “
( a + 3)
a(a + 5) ” e + 1 ’
0 - 5
a + 5 _
- 5
a + 5
(f l- 5 ) a
<3
+ 5
1)
2)
<2
+ 3
я
2
+3д
fl + 3
д (д + 3)
(я + 3)д
я (а + 3)
a2 - 5a - a - 5
_ a
2
-
6a
- 5
д(я + 3)
д(д + 3)
9 - S. Alixonov
129
2-misol. (
x 2 - 5xy
x 2 + 5xy
x 2 + у2
Y e c h i s h :
| \
------ ------- — ------ • • ---- ^
‘ --------
} x 1 - 5xy
x 2 + 5xy
x( x - 5y)
x( x + 5^)
5x - y
5x - y _ 5х + у
5 j c - j
(5x + y )(x + 5y) |
( 5 x - y ) ( x - 5 y )
x( x - 5 j)(x + 5j>)
x{x + 5y)(x ~ 5y)
_ 5x2 +xy + 25xy + 5у2 + 5хъ~ xy - 25xy + 5y2
x{x + 5)(x - 5y)
10x2 + lOy2
x(x + 5 y )(x - 5 y ) ’
10x2 - 10y2
x2 ~25,y2 _*4
2) x(x + 5 y )(x - 5 y )
x 2 + y 2
ю (х 2 +)>2)
( x - 5 ) ( x + 5y) _
10
x(x + 5y)(x - 5 у)
x 2 + у2
X
M U STAQ IL YECHISH UCHUN M ISO LLAR
1. (2x + 1-----: (2x -
— ).
Javobi: ~2x.
1 -2 x
2л: - I
Javobi: ----- .
1 - a
Javobi: 1,5.x.
Javobi: ---- - .
a + 2
5. ( j 2 ‘ 4) (
У + 2
y - 2
Javobi:
y = l
130
„
t
,
2ab . .a - b
b4
6. (a + b ------- - ) . (------ + —).
Javobi:
a.
a + b
a + b
a
1
wi2
1. (m + 1 — — -— ) : ( m ------ ).
Javobi: -m.
1 - m
m -1
8
.
x - 2 у
1
x + 2y
x + 2y
x2 - A y 2 (2y - x ) 3
(x + 2yy
x 2 - 2xy
4y2
'
Javobi:
9. ( 2a _
I 0!...... , : (
? «
+ _ L ). Javobi: 2 £ * z i 5 l
2a + b
4a + 4ab + b
4a - b
b -2 a
2a+ b
a - 1
1 - 3a + a2
1
3a + ( a - l ) 2 ~
a3 - l
~~a~-\
10
.
a2 - I
T - д
Javobi:
a +1
(a2 +a + \)(a2 -1)
/ 1
3
3
ч /
2x - L
H<(— Г----з— 7 +
~2
------- r) (•*------- r)-
Javobi: I.
X + l
XJ +1
X — X + 1
X + l
i', /
ть
. ,
lab
f
2a
12, (a + 2b+
(a ------ +
Javobi:
a - 2 b
a + 2b
a - 2 b '
13. a2 - b 2 - c 2 + 2be : ---- -— :.
Javobi: a2 ~ (b + c)2.
a + b + c
_ /5x2 - 15xy
3xy + 9y2
4 , /5
34
xy
14. ^
2
л 2
“ 1
^
I T T ' ‘
Javobi:
r - -
x + 9 jr
* + 6xy + 9y
У
x
x + 3y
,
4a2 - 6ac
_
бде + 9c2
.
6a + 9c
3
4a2 - 6ac + 9c2
4a2 + 6ac + 9c2
4a2 + 9c2 ^avo^- 2 a - 3 c '
16. (* -
+ -У) ’ (x +
~
Javobi: x2 — у2.
17. (a -
+1) : (1 - ~ —).
Javobi: —a.
1 - a
1- a
131
18. <*b +
ab
a + b
a + b
a - b
- a - b
Javobi:
ab
гУ ~xy
19. ( ^ - ^ - - x y + y < ) ....~--- +
20
.
x - y
x + y
a - b
Javobi:
—xy— 1.
(2a - b)1
4a2 - b2
(2a + b)
4 a2 + 4ab + b2
16 a
21./" 0\2 '
( c - 2 )
с + 2У
( с - 2)2
с2 - 4
22
.
25
я + 5a + 25
5 - а
а -125
Javobi:
2 ■
Javobi:
( 2а- b )
4(с + 2):
Зс2 +4
2о
2о3 + 10аМ ( ъ . 13f l -a2 - 3 0 4
д - 5 +
Javobi:
25
а2 + 5а + 25
3-§. Irratsional ifodalarni ayniy almashtirish
Agar berilgan matematik ifodada irratsional ifoda qatnashgan
bo‘ lsa, ayniy almashtirishlar orqali irratsional ifodani ratsional ifoda
ko‘ rinishga keltiriladi va u hisoblanadi. Irratsional ifoda bu ildizlardan
yoki butun son b o ‘ lmagan ratsional ko‘ rsatkichli darajadan tashkil
topgan algebraik ifodadir. Shuning uchun irratsional ifodaga quyidagicha
ta’ rif berilgan.
T a ’ rif. Agar berilgan algebraik ifodada ildiz chiqarish amali
qatnashsa, bunday ifoda irratsional ifoda deyiladi.
Irratsional ifodalarni ayniy almashtirish orqali ratsional ifoda
ko‘ rinishiga keltirish uchun asosan ildiz ostida qatnashayotgan birhad
yoki ko‘ phadni ildiz ostidan chiqarish, imkoniyati boricha maxrajni
irratsionallikdan qutqarish, nom a’ lum o ‘ zgaruvchilar kiritish orqali
berilgan irratsional ifodani ratsional ifoda ko‘ rinishiga keltirish kabi
ishlar qilinadi.
132
Bundan tashqari, o'quvchilarga sonning arifmetik ildizi va uning
kvadrat ildizi hamda irratsional ifodalarning xossalari kabi tushunchalar
tushuntirib o ‘tilib, so‘ngra quyidagi ko‘rinishdagi misollami yechish
maqsadga muvofiqdir.
1-misol.
1
1
л/5-л/З
V3-л/5
Y e c h i s h .
1
ni hisoblang.
1
1
+
■ J s -y f i
-Уз->/5
V 5-V 3
- J s S
S
- Л
2 ( ^ + ^ )
Ц -Я + Л )
j -
к
~ ( S - S ) { S + S ) ~
• •
5 - 3
2-misol.
4a
3 a V ?
4a*
2 +
4
~Ja
■\-ц
ifodani soddalashtiring.
Y e c h i s h .
n
—
3a4a* _ 4a* _ 4a
3
a^la4a
_
4a* a _
2
4
4a
2
4
4
a
4a
3
аъ4а
a14a
24а+Ъаъ4a - a4a
4a
- 3
.
----------- ------------ = ~
- e*
2)
^ - ( 2 + За3 - a ) : (-Т а ) = - ^ ( 2 + За2 - a).
3- misol. 4a* + a49a
-
— 4a*
ifodani soddalashtiring.
a
Y e c h i s h .
4a* + a49a
-
^
4a* - a4a
+ 3 j л/ot
- aVa
=
Зал/д.
^—ryj(a2 - 2ab
+ b7 )(a2 - b2 )(a + b) •
4-misol. Q_ ij
ifodani soddalashtiring.
аъ - Ь ъ
И(° + ЬУ
133
Y ech ish .
—
t](a2
- 2
ab + b2)(a2 - b2)(a
+
b)
a - b
a3
- b 3
tl(a + b)2
=
- Ж а - b)2(a
+
b)(a
+ Й) • - ^ L j L
a " 6
^/(a + Z»)2
=
( a - b ) -
yj(a
+ b)2 ■
,g л.
Ь
= a (a 3 - 63).
a ~ b
m + b ) 2
5-m isol.
xVx
+ y*Jx
2
y/у
\ X + y \ X
I
-----
,
ч
2 - s j y
r-
/—
• (,* У/ + (—
i—
ifodani sodda-
V x
+ s]y
V x + V J
lashtiring.
Y e c h ish .
xVx
+
yVx
у—
_ xVx
+>>Ух -
^Jxy%
■yfx - J x y - y f y
_
1}
V J+V ?
^ + 7 ?
_ X-s/x + У Vx - x-y/j - у Vx _
X {J* ~ S ) '
Vx +
yfy
j x + yfy
x ( J x - J y ) t
4_ x ( V x + ^ )
j
2)
- Б + f i
Л х ~ у ) ’°
< Л +ч6 о < Л - 7 7 >
( л + Я
я
2^у
* +
2 ' Iy ( ‘Jx + J y )
_
x + 2 jx y
+2у
3)
( л + ^ ) 2 + ^ +
^ '
( v j + v ? ) 2
= ( л ; ^ ) 2
'
1.
Kasrli irratsional ifodalarning maxrajlarini berilishiga qarab irratsio-
nallikdan quyidagicha chiqariladi.
y/a±\fb
ko'rinishlarda berilgan bo'lsa, ularning o‘zaro ко‘paytmasi
(Va ±
-Jb)(Ja - 4b)
= a - b
bo'ladi. Agar irratsional ifodalar ^ + ^ va
134
А
j 2
—
ko'rinishlarda berilgan bo'lsa, ularning maxrajlari irratsionallikdan
quyidagicha chiqariladi:
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