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Qisqa ko’paytirish formulalarining 3-dan foydalanib, ifodani ko’paytuvchilarga ajratamiz: (2x − 2 + x + 3)(2x − 2 − x − 3) ≤ 0 ⇐⇒ (3x + 1)(x − 5) ≤ 0. Bu tengsizlikni oraliqlar usuli bilan yechib, [− 1 3 ; 5] yechimni olamiz. Bu kesmada 0, 1, 2, 3, 4, 5 butun sonlari yotadi, ular 6 ta. Javob: 6 (A). 11. (01-12-17) Tengsizlikni yeching. |x + 1| > 2|x + 2| A) (−2; −1) B) [−3; −1] C) (−3; − 5 3 ) D) (−3; 0) 12. (03-7-22) Tengsizlikni yeching. |x − 4| < |x + 4| A) (−4; 4) B) (0; 4) ∪ (4; ∞) C) (0; ∞) D) (−∞; −4) ∪ (−4; 0) 13. (01-5-24) Tengsizlikni butun yechimlari yig’indisini toping. |5 − 2x| ≤ 3 A) 10 B) 15 C) 6 D) 3 84 14. (02-1-47) Tengsizlikni qanoatlantiruvchi butun son- lar nechta? |3x + 8| ≤ 2 A) 1 B) 2 C) 3 D) 4 15. (96-3-26) |x − 1| ≥ 2 tengsizlikni yeching. A) (−∞; −1] B) [−1; 3] C) (−∞; −1] ∪ [3; ∞) D) [1; 3] Yechish: Misolni 3-usul, ya’ni geometrik usul- dan foydalanib yechamiz. |x − 1| miqdor son- lar o’qida x va 1 nuqtalar orasidagi masofani ifo- dalaydi. Demak, berilgan tengsizlikning yechimi sonlar o’qida koordinatasi 1 bo’lgan nuqtadan ma- sofasi 2 birlik va undan katta x ning barcha qiy- matlaridan iborat (6.3-chizmaga qarang). Son- lar o’qida koordinatasi 1 bo’lgan nuqtadan 2 bir- lik chapda −1 ni, 2 birlik o’ngda yotuvchi 3 ni, topamiz. Shunday qilib tengsizlikning yechimi (−1; 3) intervalning tashqarisi bo’ladi. Demak, yechim (−∞; −1] ∪ [3; ∞) dan iborat. Javob: (−∞; −1] ∪ [3; ∞). (C) 16. (96-7-8) Tengsizlik nechta butun yechimga ega? |x − 2| ≤ 5 A) 11 B) 10 C) 8 D) 7 17. (96-11-27) Tengsizlikni yeching. |x − 1| ≥ 1 A) [0; 2] B) (−∞; 0] ∪ [2; ∞) C) [−2; 0] D) [2; ∞) 18. (96-12-27) Tengsizlikni yeching. |x − 1| ≤ 2 A) (−∞; 3] B) (−∞; −1] ∪ [3; ∞) C) [−1; 3] D) [1; 3] 19. (97-7-8) Tengsizlik nechta butun yechimga ega. |x + 2| ≤ 3 A) 5 B) 6 C) 7 D) 4 20. (97-1-73) Tengsizlikning eng katta natural yechi- mini toping. |3x − 7| < 5 A) 4 B) 3 C) 2 D) 1 21. (97-3-8) Tengsizlik nechta butun yechimga ega. |3 − x| < 4 A) 4 B) 5 C) 6 D) 7 22. (97-10-8) Tengsizlik nechta butun yechimga ega? |4 − x| < 6 A) 3 B) 5 C) 8 D) 11 23. (98-5-23) Tengsizlikning eng kichik natural yechi- mini toping. |x − 7| ≤ 1 A) 5 B) 7 C) 8 D) 6 24. (99-7-24) Tengsizlikning eng kichik natural yechi- mini toping. |x − 6| ≤ 8 A) 2 B) 7 C) 3 D) 1 25. (99-9-18) Ushbu |x − 4| ≤ 12 tengsizlikning eng kichik va eng katta butun yechimlari yig’indisini toping. A) 6 B) 8 C) −6 D) −8 6.3 Modulli tenglamalar va tengsizlik- lar sistemasi 1. (98-8-25) b ning qanday qiymatlarida tenglamalar sistemasi yagona yechimga ega? ½ x = 3 − |y| 2x − |y| = b A) b = 0 B) b > 0 C) b < 1 D) b = 6 Yechish: Agar (x 0 ; y 0 ) berilgan sistemaning ye- chimi bo’lsa, u holda |y| = | − y| tenglikka ko’ra, (x 0 ; −y 0 ) ham uning yechimi bo’ladi. Demak, sistema yagona yechimga ega bo’lishi uchun y 0 = 0 bo’lishi kerak ekan. Uni sistemaning birinchi tenglamasiga qo’yib x = 3 ekanini, ikkinchi tengla- madan esa b = 6 ekanini hosil qilamiz. Javob: b = 6 (D). 2. (98-1-25) a ning qanday qiymatlarida ½ 3|x| + y = 2 |x| + 2y = a sistema yagona yechimga ega? A) a = 0 B) a > 0 C) a = 2 D) a = 4 3. (98-8-23) Agar ½ x + 2|y| = 3 x − 3y = 5 bo’lsa, x − y ning qiymatini toping. A) 3 B) 2 C) 1 D) −1 4. (98-1-23) Agar ½ |x| + y = 2 3x + y = 4 bo’lsa, x + y ning qiymatini toping. A) 3 B) 1 C) 2, 5 D) 2 85 5. (00-1-18) Tengsizliklar sistemasini yeching. ½ x ≥ 3 |x − 3| ≤ 1 A) 2 ≤ x ≤ 3 B) −2 ≤ x ≤ 4 C) 3 ≤ x ≤ 4 D) x ≤ 4 Yechish: 6.2-dagi 1-qoidaga ko’ra berilgan sis- temaning 2-tengsizligi −1 ≤ x − 3 ≤ 1 qo’sh tengsizlikka teng kichli. Bu tengsizlikning har bir qismiga 3 ni qo’shib 2 ≤ x ≤ 4 ni olamiz. Sistemaning 1-tengsizligi x ≥ 3 ni hisobga olsak, 3 ≤ x ≤ 4 javobni olamiz. Javob: 3 ≤ x ≤ 4 (C). 6. (02-10-55) Tengsizliklar sistemasini yeching. ½ |2x − 3| ≤ 1 5 − 0, 4x > 0 A) [1; 2] B) (−∞; 2] C) (−∞; 1] ∪ (2; ∞) D) (−0, 4; 2) 7. (01-7-20) Tenglamalar sistemasi nechta yechimga ega? ½ |x| + |y| = 1 x 2 + y 2 = 4 A) 1 B) 2 C) 4 D) ∅ 8. (02-12-17) Agar ½ |x − 1| + |y − 5| = 1 y = 5 + |x − 1| bo’lsa, x+y qanday qiymatlar qabul qilishi mum- kin? A) 6 yoki 8 B) 7 C) 8 yoki 10 D) 6 yoki 7 9. (00-7-20) Agar ½ (x − 2) 2 + |y| = 4 |x − 2| + |y| = 2 bo’lsa, x + y ning qiymatini toping. A) 4 yoki 2 yoki 0 B) 0 yoki 3 C) 2 yoki 4 D) 0 yoki 4 10. (03-10-31) Agar ½ |x + y| = 5 xy = 4, 75 bo’lsa, son o’qida x va y sonlari orasidagi maso- fani toping. A) √ 6 B) √ 3 C) √ 5 D) √ 7 7 -bob. Irratsional tenglama va tengsizliklar 7.1 Irratsional tenglamalar Noma’lumi ildiz belgisi ostida bo’lgan tenglamalar irratsional tenglamalar deyiladi. Masalan, 4 √ x − 3 = 3 − √ x; (5 − x) √ x − 3 = 3 − 3 √ 2x tenglamalar irratsional tenglamalardir. Irratsional teng- lamalarni yechish ma’lum bir shakl almashtirishlar yor- damida ularni ratsional tenglamalarga keltirishga asos- langan. Radikallardan (ildiz belgisidan) qutilish maqsa- dida tenglamaning ikkala qismini bir xil darajaga ko’ta- riladi. Lekin darajaga ko’targanda chet ildizlar hosil bo’lishi mumkin. Shuning uchun oxirgi tenglamani ye- chishda topilgan ildizlarni, berilgan irratsional tenglama- ning o’ziga qo’yib tekshirib ko’rish lozim. Irratsional tenglamalarni yechishda ko’p qo’llaniladi- gan quyidagi tengkuchliliklarni keltiramiz. 1. 2k p f (x) = ϕ(x) ⇐⇒ ½ f (x) = [ϕ(x)] 2k ϕ(x) ≥ 0 2. 2k+1 p f (x) = ϕ(x) ⇐⇒ f (x) = [ϕ(x)] 2k+1 Irratsional tenglamalarni yechish usullarini misol- larda namoyish qilamiz. 1-misol. √ x + 2 + x = 0 tenglamani yeching. Yechish: Tenglamada x ni tenglikning o’ng qis- miga o’tkazamiz va ikkala qismini kvadratga ko’taramiz, natijada x + 2 = (−x) 2 ⇐⇒ x 2 − x − 2 = 0 kvadrat tenglamaga kelamiz. Bu kvadrat tenglamaning ildizlari x 1 = −1 va x 2 = 2 lardir. Topilgan yechim- larni berilgan tenglamaga qo’yamiz. Dastlab x 1 = −1 ni qo’yamiz: √ −1 + 2 + (−1) = 1 − 1 = 0. Endi x 2 = 2 ni √ 2 + 2 + 2 = 2 + 2 = 4 6= 0. Demak, x 2 = 2 chet ildiz ekan, x 1 = −1 esa tenglamani qanoatlantiradi. Javob: x = −1. 2-misol. √ x + 1 = −7 tenglamani yeching. Yechish: Arifmetik ildizning manfiymasligidan √ x + 1 ≥ 0 ekanligi kelib chiqadi. Tenglamaning o’ng qismi −7 esa manfiy son. Shuning uchun tenglama ildizga ega emas. Javob: ∅. Xuddi shunday ko’rsatish mumkinki, |x 2 − 1| + √ 2x + 1 = −1 irratsional tenglama ham yechimga ega emas. 3-misol. Quyidagi tenglamani yeching: x 2 − 3x + p x 2 − 3x + 5 = 7. (1) Yechish: Tenglamada x 2 −3x = t belgilash olamiz, natijada berilgan tenglama t + √ t + 5 = 7 ko’rinishga ega bo’ladi. Bu tenglamada t ni tenglikning o’ng qis- miga o’tkazib, keyin tenglikning har ikkala qismini kvad- ratga ko’tarib, t + 5 = (7 − t) 2 kvadrat tenglamaga ke- lamiz. Bu kvadrat tenglamani yechib t 1 = 4, t 2 = 11 ni olamiz. t 2 = 11 ildiz t + √ t + 5 = 7 irratsional tenglamani qanoatlantirmaydi. Shuning uchun faqat t 1 = 4 qiymatni x 2 − 3x = t belgilashga qo’yib, x 2 − 3x = t 1 ⇐⇒ x 2 − 3x = 4 kvadrat tenglamaga ega bo’lamiz. Bu kvadrat tenglama- ning ildizlari x 1 = −1, x 2 = 4 lardir. Tekshirish mum- kinki, ular (1) tenglamani ham qanoatlantiradi. Javob: x 1 = −1, x 2 = 4. 86 1. (98-9-19) Tenglama ildizlari yig’indisini toping. p x 4 + 5x 2 = −3x A) 0 B) −2 C) −4 D) 2 Yechish: 1-qoidaga ko’ra berilgan tenglama ½ x 4 + 5x 2 = 9x 2 −3x ≥ 0 sistemaga teng kuchli. Uning birinchi tenglamasini x 4 − 4x 2 = 0 ⇐⇒ x 2 (x 2 − 4) = 0 shaklda yozib olamiz. Bu tenglamaning yechim- lari x 1 = 0, x 2 = −2, x 3 = 2 lardir. Endi −3x ≥ 0 shartni tekshiramiz. Uni faqat x 1 = 0 va x 2 = −2 sonlar qanoatlantiradi. Shunday qilib, beril- gan tenglamaning ildizlari x 1 = 0 va x 2 = −2 ekan. Ular yig’indisi 0 + (−2) = −2. Javob: −2 (B). 2. (97-1-72) Tenglamani yeching. √ x + 2 + x = 0 A) −1 B) −2 C) 2 D) 0 3. (97-5-39) Tenglamalar sistemasini yeching. ½ p (x + 2) 2 = x + 2 p (x − 2) 2 = 2 − x A) x ≥ −2 B) x < 2 C) x ≤ 2 D) −2 ≤ x ≤ 2 4. (97-7-61) Ushbu √ 3 + 2x = −x tenglik x ning qanday qiymatlarida o’rinli? A) −1 B) 1 C) −3 D) 3 5. (98-2-21) Agar √ x 4 − 9x 2 = −4x tenglamaning katta ildizi x 0 bo’lsa, x 0 + 10 nechaga teng? A) 10 B) 12 C) 20 D) 15 Yechish: Berilgan tenglamaning har ikkala tomo- nini kvadratga ko’tarib, o’ng tomondagi hadni chap tomonga o’tkazib, x 4 − 25x 2 = 0 ⇐⇒ x 2 (x 2 − 25) = 0 tenglamani olamiz. Bu tenglamaning ildizlari x 1 = −5, x 2 = 0, x 3 = 5 lardir. Lekin x 3 = 5 dastlabki tenglamani qanoatlantirmaydi. Shu- ning uchun tenglamaning katta ildizi x 2 = 0 bo’- ladi. x 0 + 10 = 0 + 10 = 10. Javob: 10 (A). 6. (99-2-19) Tenglamaning ildizlari yig’indisini to- ping. p x 2 − 3x + 5 + x 2 − 3x = 7 A) 4 B) −3 C) 3 D) −4 7. (99-6-41) √ a − √ b = 4 va a − b = 24 bo’lsa, √ a + √ b nimaga teng. A) 6 B) 4 C) 5 D) 3 8. (99-5-15) Tenglamaning natural ildizlari nechta? p (3x − 13) 2 = 13 − 3x A) 0 B) 1 C) 2 D) 4 9. (99-8-3) Tenglamani yeching. √ x + 1 + √ 2x + 3 = 1 A) −1 B) 3 C) −1; 3 D) 1 10. (99-9-11) Tenglama ildizlari o’rta arifmetigini to- ping. (x 2 − 25)(x − 3)(x − 6) √ 4 − x = 0 A) 4 1 3 B) 1 1 3 C) 2 3 D) 4 1 2 11. (99-9-12) Tenglama ildizlari ko’paytmasini toping. p x 2 + 77 − 2 4 p x 2 + 77 − 3 = 0 A) −3 B) 3 C) 4 D) −4 Yechish: Tenglamada 4 √ x 2 + 77 = t belgilash olamiz, natijada berilgan tenglama t 2 −2t−3 = 0 ko’rinishga ega bo’ladi. Bu kvadrat tenglama yechimlari t 1 = −1, t 2 = 3 lardir. Yechimlarni 4 √ x 2 + 77 = t belgilashga qo’yib, 4 √ x 2 + 77 = −1 va 4 √ x 2 + 77 = 3 tenglamalarni hosil qilamiz. Ammo 4 √ x 2 + 77 = −1 tenglama yechimga ega emas (chunki tenglamaning chap qismi musbat, o’ng qismi esa manfiy). Ikkinchi tenglamaning har ikkala qismini 4-darajaga ko’tarib, x 2 + 77 = 3 4 ⇐⇒ x 2 = 4 tenglamaga ega bo’lamiz. Bu tenglamaning ye- chimlari x 1 = −2, x 2 = 2 lardir. Tekshirish nati- jasi ko’rsatadiki, x 1 = −2, x 2 = 2 lar dastlabki tenglamani qanoatlantiradi. Ularning ko’paytmasi (−2) · 2 = −4. Javob: −4 (D). 12. (00-1-19) Agar r 1 − 1 x = x − 1 x − 6 bo’lsa, 6 1 8 + x ning qiymatini hisoblang. A) −7 B) 6 C) 7 D) −6 13. (00-2-19) Tenglamani yeching. p (2x − 1) 2 (3 − x) = (2x − 1) √ 3 − x A) [0, 5; 3] B) [0; 3] C) [1; 3] D) (−∞; 0, 5] Yechish: Tenglama x ≤ 3 da ma’noga ega. De- mak, berilgan tenglama ½ p (2x − 1) 2 (3 − x) = (2x − 1) √ 3 − x x ≤ 3 87 sistemaga teng kuchli. Uning birinchi tenglamasi x ≤ 3 shartda |2x − 1| p (3 − x) − (2x − 1) √ 3 − x = 0 tenglamaga teng kuchli. Bu tenglama esa √ 3 − x · (|2x − 1| − (2x − 1)) = 0 ga teng kuchli. Bu tenglama yechimlari esa √ 3 − x = 0 hamda |2x − 1| = 2x − 1 tenglama yechimlari bilan ustma-ust tushadi. 1-tenglamadan x = 3 ni olamiz. 2-tenglik modul ta’rufiga ko’ra (6.1-ning 1-qoidasiga qarang) 2x − 1 ≥ 0 shartda o’rinli. Bu tengsizlikning yechimi x ≥ 0, 5 dan iborat. Agar x ≤ 3 shartni hisobga olsak, x ∈ [0, 5 ; 3] ni olamiz. Javob: [0, 5 ; 3] (A). 14. (00-3-10) Tenglamani yeching. 3 √ 2x − 5 √ 8x + 7 √ 18x = 28 A) 1 B) 2 C) 3 D) 4 15. (00-3-22) Tenglama ildizlari yig’indisini toping. √ x + 1 + √ 2x + 3 = 1 A) 2 B) 3 C) 4 D) −1 16. (00-4-7) Tenglamani yeching. 2 √ x − √ 2x 2 + 3 = √ x + 1 A) 8 B) 4 C) 9 D) 1 17. (00-5-29) Tenglamani yeching. p x 2 − x − 2 = x − 3 A) 5 B) 2,2 C) 4 D) ∅ 18. (00-6-33) Agar p 3x 2 − 6x + 16 = 2x − 1 bo’lsa, x 2 · (x + 2) ning qiymatini toping. A) −75 B) −45 C) 15 D) 45 19. (00-8-5) Tenglamani yeching. (x 2 − 9) √ x + 1 = 0 A) −1; 3 B) ±3 C) ±3; 1 D) 2 20. (00-8-25) Agar √ 8 − a + √ 5 + a = 5 bo’lsa, p (8 − a)(5 + a) ning qiymatini toping. A) 6 B) 20 C) 12 D) 10 Yechish: Tenglamaning ikkala qismini kvadratga ko’taramiz, natijada 8 − a + 2 √ 8 − a √ 5 + a + a + 5 = 25 ⇐⇒ 2 p (8 − a)(5 + a) = 12 tenglikni olamiz. Bu yerdan p (8 − a)(5 + a) = 6 ni olamiz. Javob: 6 (A). 21. (00-8-26) Agar p 25 − x 2 + p 15 − x 2 = 5 bo’lsa, √ 25 − x 2 − √ 15 − x 2 ifodaning qiymatini toping. A) 2 B) 3 C) 5 D) 6 22. (00-9-31) Agar 4 √ ab = 2 √ 3 va a, b ∈ N bo’lsa, a− b quyidagi keltirilgan qiymatlardan qaysi birini qabul qila olmaydi? A) −32 B) 10 C) 0 D) 25 23. (97-5-26) Agar ½ √ x + √ y = 3 √ xy = 2 bo’lsa, x + y ni toping. A) 2 B) 3 C) 4 D) 5 24. (01-1-19) Agar ½ x − y = 21 √ x − √ y = 3 bo’lsa, x + y ning qiymatini toping. A) 7 B) 12 C) 23 D) 29 25. (01-2-24) Tenglama ildizlari o’rta arifmetigini to- ping. x − 5 √ x + 4 = 0 A) 16 B) 8, 5 C) 3 D) 2 26. (01-5-9) Tenglama ildizlari yig’indisini toping. (x 2 − 4) √ x + 1 = 0 A) 1 B) −1 C) 3 D) 2 Yechish: Tenglamaning chap qismi x ≥ −1 da ma’noga ega. Ko’paytma nolga aylanishi uchun, ko’paytuvchilardan biri nolga aylanishi yetarli. Shuning uchun x 2 − 4 = 0 va x + 1 = 0 tenglama yecimlari ichidan x ≥ −1 shartni qanoatlantiru- vchilarini olishimiz kerak. Ularning yechimlari x 1 = −2, x 2 = 2 va x 3 = −1 lardir. x 1 = −2 yechim x ≥ −1 shartni qanoatlantirmaydi. De- mak, x 2 = 2 va x 3 = −1 lar berilgan tenglama- ning yechimi bo’ladi. Ularning yig’indisi 2+(−1) = 1. Javob: 1 (A). 27. (01-6-25) Agar √ x + 1 + x − 11 = 0 bo’lsa, x + 12 ning qiymatini toping. A) 15 B) 16 C) 20 D) 19 28. (01-9-12) Ushbu ½ y = √ 16 − x 2 y − x = 4 tenglamalar sistemasining nechta yechimi bor? A) 2 B) 1 C) 0 D) 3 88 29. (01-10-20) Tenglama nechta ildizga ega? x − 9 √ x + 3 = x − 15 A) 0 B) 1 C) 2 D) 3 30. (01-12-43) Tenglamani yeching. √ 3x − 7 − √ 7 − 3x = 0 A) 2, 3 B) 3 7 C) 7 3 D) ∅ 31. (02-1-8) Tenglama nechta ildizga ega? p x 2 + 1 − p x 2 − 1 = 1 A) ∅ B) 1 C) 2 D) 3 32. (03-7-20) Tenglamani yeching. 3 r x 3 q x 3 √ x... = 8 A) 56 B) 48 C) 60 D) 64 33. (03-8-38) Tenglamani yeching. √ x + 4 √ x − 12 = 0 A) 81 B) 16 C) 25 D) 9 7.2 Irratsional tengsizliklar Irratsional tengsizliklarni yechishda quyidagi teng kuch- liliklardan foydalaniladi. 1. 2 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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