metodlaridan foydalanamiz. Misol: 1)
tenglik xossasidan foydalanib
ko`rinishga keltiramiz.
.
2)
avval ratsional
ifoda xossasidan foydalanamiz,
ildiz xossalaridan ham foydalanamiz.;
deb tenglikning ikkala tarafini
ga
ko`paytiramiz,
soddalashtiramiz,
;
tenglikning
ikkala
tarafini
kvadratga
oshiramiz
tenglik xossasidan foydalanib, 5x
2
+4x
2
+8x+6x-
27+4=0; 9x
2
+14x-23
=
0 kvadrat
tenglama ildizlarini
topish formulasidan
foydalanamiz. Ikkinchi had juft bo`lgani uchun ikkiga bo`lib yuboramiz.
Diskriminant D
=
49+9*23
=
256 x
1
=
-29/3 x
2
=
1? tenglamani yechgandan keyin uni
albatta tekshirishimiz shart. Ya`ni topilgan yechimni tenglamaga qo`yib, to`g’ri
tenglik bajarilish yoki bajarilmasligini analiz qilamiz.
1)
+4/
=
2
;
/3+4*3/
=
2/3; 14/3+4*3/2
=
2/3*2; 14/3+6
=
4/3;
2)
+4/
=
2
; 2+2
=
2*2;
Demak topilgan yechim har doim ham tenglamani qanoatlantirmas ekan. J:
x
=
1. Ayrim hollarda ildiz ostidagi
bajarilish shartlarini, ya`ni tenglamadagi
noma`lumning qabul qila oladigan qiymatlarini tekshirsak ham bo`ladi. Ya`ni
topilgan yechim shu oraliqqa tegishlimi yoki yo`qmi? Yuqoridagi tenglamada bu
oraliq
3+x > 0 x > -3
9-5x => 0 x <
=
9/5 -3 < x < 9/5
topilgan ikkala ildiz ham shu oraliqqa tegishli. Lekin
ulardan biri tenglamani
qanoatlantirmayapdi. Demak albatta topilgan yechimni tenglamaga qo`yib
tekshirish kerak ekan.
3)
х
х
12
5
12
5
х
х
12
1
5
х
1
5
12
х
х
х
х
3
2
3
4
5
9
0
3
х
х
3
4
2
6
3
5
9
х
х
х
2
2
5
6
27
2
х
х
х
4
8
2
4
2
5
6
27
x
x
x
x
9
/
23
*
5
9
9
/
23
3
9
/
23
3
196
23
27
1
*
5
9
1
3
1
3
;
3
5
1
3
3
5
x
x
x
5
3
5
3
0
3
5
5
3
0
3
5
x
x
x
x
x
;
1
3
3
5
3
5
x
x
x
;
1
3
9
25
2
2
2
x
x
1
6
9
9
25
2
2
x
x
x
;
2
:
/
0
10
6
2
16
x
x
- aniqlanish sohaga kirmaydi.
Tek:
Viyet teoremasidan foydalanib tanlash yo`li bilan ham ildizlarni topish
mumkin. x
2
-8x-9=0; x
1
+x
2
=8, x
1
*x
2
=-9
Agar x
1
=-1, x
2
=9 desak, Viyet teoremasi sharti bajariladi.
O’zgaruvchi kiritish usuli. 1) x
4
-4x
2
-5=0; x
2
=z
z
2
-4z-5=0;
z
1
+z
2
=4;
z
1
*z
2
=-5 demak, z
1
=-1, z
2
=5, z
1
-chet ildiz x
2
=5; x
1, 2
=
tenglamaning
yechimi.
2)
- chet ildiz
1)
2)
8. x
4
+6x
3
+5x
2
-12x+3=0
x
4
+6x
3
+9x
2
-4x
2
-12x+3=0
(x
2
+3x)
2
-4(x
2
+3x)+3=0
x
2
+3x=z; z
2
-4z+3=0
Tanlash yo`li
bilan
1)
2)
0
5
3
8
2
x
x
2
13
169
5
8
4
9
D
;
8
5
16
10
16
13
3
1
x
.
1
16
13
3
2
x
1
x
;
3
1
5
1
1
3
3
1
5
;
2
4
8
2
2
2
2
5
;
2
3
1
1
2
2
x
x
x
x
,
1
x
0
x
;
1
2
z
x
x
;
2
3
1
z
z
0
2
3
2
2
z
z
;
25
2
2
4
9
D
;
2
1
4
5
3
1
z
2
4
5
3
2
z
1
z
;
2
1
2
x
x
2
1
x
x
;
2
2
x
x
;
2
2
1
x
;
2
2
2
1
2
1
2
2
1
2
1
x
2
2
1
x
;
2
1
x
x
;
2
2
x
x
2
2
1
x
;
2
2
2
1
2
1
2
2
1
2
x
2
2
2
x
,
1
1
z
3
2
z
;
1
3
2
x
x
,
0
1
3
2
x
x
13
4
9
D
;
2
13
3
1
x
;
2
13
3
2
x
;
3
3
2
x
x
,
0
3
3
2
x
x
21
12
9
D
9.
Tenglamaning har ikkala tomonini
ga bo`lamiz.
bundan
desak,
1)
2)
Modul qatnashgan tenglamalar
Modul tushunchasi matematikaning muhim tushunchalaridan biri hisoblanadi.
Ta’rif.
sonining moduli deb, agar u son nomanfiy bo’lsa, sonning o’ziga
agar u son nomanfiy bo’lsa,
soniga aytiladi.
Ta’rifga ko’ra,
o’zgaruvchisi modul ostida qatnashgan tenglamalarni
yechish uchun quyidagi
metodlardan foydalaniladi.
1) Ta’rifdan foydalanib yechiladi.
2) Tenglamani ikkala tomonini kvadratga oshiriladi.
3) Oraliqlarga ajratib yechiladi.
3-misol.
yeching.
Yechish.
1-usul. Ta’rifga ko’ra
Birinchi sistemadan
yechimga keltiramiz.
Ikkinchi sistemadan
2-usul. Tenglikni ikkala tomonini kvadratga oshiramiz.
;
2
21
3
3
x
2
21
3
4
x
0
2
3
16
3
2
2
3
2
x
x
x
x
0
2
x
0
2
3
16
3
2
2
2
x
x
x
x
0
16
1
3
)
1
(
2
2
2
x
x
x
x
z
x
x
1
2
1
2
2
2
z
x
x
;
0
16
3
2
2
2
z
z
0
20
3
2
2
z
z
;
169
20
2
4
9
D
;
4
4
13
3
1
z
2
5
4
10
2
z
;
4
1
x
x
;
0
1
4
2
x
x
.
3
1
4
D
;
3
2
1
x
;
3
2
2
x
;
2
5
1
x
x
;
0
2
5
2
2
x
x
.
9
2
2
4
25
D
;
4
1
4
3
5
3
x
2
4
3
5
4
x
a
a
a
lsa,
bo'
0
a
agar
a,
-
lsa,
bo'
0
a
agar
,
a
a
3
1
4
x
3
1
4
0
1
4
x
x
3
1
4
0
1
4
x
x
1
1
x
2
1
2
x
