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MODULLI TENGLAMA VA TENGSIZLIKLARNI YECHISH USULLARI

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20.11.2024

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Matematika va informatikani o’qitish texnologiyalari loyihalari MAVZU: MODULLI TENGLAMA VA TENGSIZLIKLARNI YECHISH USULLARI i9 presentation to Joe Smith2 Sonning moduli Berilgan a haqiqiy sonning moduli - sonlar o‘qining markazi yoki koordinatalar boshidan berilgan songacha bo‘lgan masofaga. ∙ ∙-5 50 5 birlik 5 birlik I 5 I = 5, 5 > 0 -5, -5 < 0 i9 presentation to Joe Smith3 Modulli chiziqli tenglamalar: Ix +4 I= 9 a) x+4=9 x=9-4 x= 5 b) x+4=-9 x=-9-4 x=-13 Javob: = 5 ; = -13 i9 presentation to Joe Smith4 |x -1 |=| 2 x +1 | (x -1 ) 2 =( 2 x +1 ) 2 (x -1 ) 2 -( 2 x +1 ) 2 =0 ((x -1 )-( 2 x +1 ))((x -1 )+( 2 x +1 ))=0 - x -2 =0 3 x= 0 -x= 2 x= 0:3 x= -2 x= 0 Javob : -2 ; 0 Modulli tenglama i9 presentation to Joe Smith5 Modulli tenglama Tenglamani yeching: | х + 2 | = | 2х - 6 | 1 - usul : х + 2 = 2х – 6 va х + 2 = - (2х – 6) х = 8 3х = 4 х = 4/3 2 - usul : ( | х + 2 | ) 2 = ( | 2х - 6 | ) 2 | а | 2 =а 2 (х + 2) 2 = (2х - 6) 2 3х 2 – 28х + 32 = 0 => х = 8, х = 4/3 i9 presentation to Joe Smith6 I x I ≤ a, bunda a > 0Noma’lum modul belgisi ostida qatnashgan tengsizliklar ∙ ∙ -a a0 I x I ≤ a [ - a; a] kesma –a ≤ x ≤ a tengsizlikni qanoatlantiruvchi x sonlar to‘plami. I x I ≤ a , tengsizlik –a ≤ x ≤ a qo‘sh tengsizlikning ayni o‘zini bildiradi. i9 presentation to Joe Smith7 ∙ - 12 12 01 ) IxI < 12 Javob: (- 12 ; 12 ) № 271 . Tengsizlikning yechimlari to‘plamini son o‘qida tasvirlang: – 12 < x < 12 3) IxI ≤ 9 ∙ - 9 9 0 Javob: [- 9 ; 9 ] i9 presentation to Joe Smith8 TENGSIZLIKNI YECHING0 9 2   x 3 3 :    х Javob 9 2  x х х  2 3  х - 3 3 0 9 2   x 3 3 :    х Javob 9 2  x х х  2 3  х i9 presentation to Joe Smith9 -2 0,8 I5x+3I < 7x ning tengsizlik bajaradigan barcha butun qiymatlarini toping – 7 < 5x+3 < 7 1) 5x+3 < 7 5x<4 x<0,8 Javob: -1 va 0 2) 5x+3> -7 5x>-10 x > -2 -1 0

Matematika va informatikani o’qitish texnologiyalari loyihalari MAVZU: MODULLI TENGLAMA VA TENGSIZLIKLARNI YECHISH USULLARI

i9 presentation to Joe Smith2 Sonning moduli Berilgan a haqiqiy sonning moduli - sonlar o‘qining markazi yoki koordinatalar boshidan berilgan songacha bo‘lgan masofaga. ∙ ∙-5 50 5 birlik 5 birlik I 5 I = 5, 5 > 0 -5, -5 < 0

i9 presentation to Joe Smith3 Modulli chiziqli tenglamalar: Ix +4 I= 9 a) x+4=9 x=9-4 x= 5 b) x+4=-9 x=-9-4 x=-13 Javob: = 5 ; = -13

i9 presentation to Joe Smith4 |x -1 |=| 2 x +1 | (x -1 ) 2 =( 2 x +1 ) 2 (x -1 ) 2 -( 2 x +1 ) 2 =0 ((x -1 )-( 2 x +1 ))((x -1 )+( 2 x +1 ))=0 - x -2 =0 3 x= 0 -x= 2 x= 0:3 x= -2 x= 0 Javob : -2 ; 0 Modulli tenglama

i9 presentation to Joe Smith5 Modulli tenglama Tenglamani yeching: | х + 2 | = | 2х - 6 | 1 - usul : х + 2 = 2х – 6 va х + 2 = - (2х – 6) х = 8 3х = 4 х = 4/3 2 - usul : ( | х + 2 | ) 2 = ( | 2х - 6 | ) 2 | а | 2 =а 2 (х + 2) 2 = (2х - 6) 2 3х 2 – 28х + 32 = 0 => х = 8, х = 4/3

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