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ax + b > 0,
ax + b ≥ 0 (5.1) ax + b < 0, ax + b ≤ 0 (5.2) ko’rinishdagi yoki soddalashtirishdan so’ng shunday ko’- rinishga keladigan tengsizlik chiziqli tengsizlik deyiladi. (5.1) va (5.2) da x noma’lum, a va b lar haqiqiy sonlar, a noma’lum oldidagi koeffitsiyent, b ozod had deyiladi. Eng sodda x > a, x ≤ b yoki a < x ≤ b, a ≤ x ≤ b ko’rinishdagi tengsizliklarning yechimlari (a; ∞), (−∞; b] yoki (a; b], [a; b] shaklda yozish qabul qilin- gan. 1. 3x + 1 < 10 tengsizlikni yeching. A) (3; ∞) B) (1; ∞) C) (−∞; 3) D) ∅ Yechish: Berilgan tengsizlikda 1 ni tengsizlik- ning o’ng qismiga o’tkazib, keyin tengsizlikning ikkala qismiga 3 ga bo’lib x < 3 ni olamiz. Javob: (−∞; 3) (C) . 2. 7x + 5 > 19 tengsizlikni yeching. A) (2; ∞) B) (7; ∞) C) (−∞; 2) D) ∅ 3. 2x + 7 ≤ 11 tengsizlikni yeching. A) [2; ∞) B) (7; ∞) C) (−∞; 2] D) ∅ 4. 5x + 9 ≥ 14 tengsizlikni yeching. A) [1; ∞) B) [5; ∞) C) (−∞; 1] D) ∅ 5. (00-6-10) Tengsizlikni yeching. 1 − 17 − 3x 2 > 1, 5x A) (−2, 5; 0) B) (−∞; −2, 5) C) (−∞; 0) D) ∅ Yechish: Berilgan tengsizlikning har ikkala qis- mini 2 ga ko’paytirsak, tengsizlik saqlanadi 2 − 17 + 3x > 3x ⇐⇒ −15 > 0 · x. Bu tengsizlik yechimga ega emas. Javob: ∅ (D). 6. Tengsizlikni yeching. 1 − x 2 < 2x + 1 4 A) ( 3 4 ; ∞) B) (1; ∞) C) (−∞; 3 4 ) D) ∅ 7. (01-8-10) Tengsizlikni yeching. 1 − x 2 + 3 < 3x − 2x + 1 4 A) (1 1 3 ; ∞) B) (1 1 13 ; ∞) C) (1 1 4 ; ∞) D) ∅ 8. (98-2-17) Quyida keltirilgan tengsizliklardan qaysi biri 3x − a > b − 2x tengsizlikka teng kuchli emas? A) 5x − a > b B) 6x − 2a > 2b − 4x C) 3x > a + b − 2x D) a − 3x > 2x − b 71 9. 2x + 5 > a ga teng kuchli tengsizlikni toping. A) 2x + 5 − a ≥ 0 B) 2x + 5 − a > 0 C) 2x < a − 5 D) −2x < −5 − a 10. (00-8-33) k ning 4y 2 −3y+k = 0 tenglama haqiqiy ildizlarga ega bo’lmaydigan eng kichik butun qiymatini toping. A) 1 B) 3 C) 4 D) 5 Yechish: 4y 2 − 3y + k = 0 kvadrat tenglama haqiqiy ildizlarga ega bo’lmasligi uchun D < 0 bo’lishi kerak: D = 9 − 16k < 0 ⇐⇒ k > 9 16 . Bu tengsizlikni qanoatlantiruvchi eng kichik bu- tun son k = 1. Javob: 1 (A) . 11. (02-2-10) Tengsizlikni qanoatlantiruvchi eng katta butun sonni toping. (x + 1) 2 > (x + 2) 2 A) −2 B) −1 C) −3 D) −4 12. (02-3-18) Tengsizlikni qanoatlantiruvchi eng kichik butun manfiy son nechaga teng? 8 + 6x − 8 10 > x − 2 6 + 1 − 5x 8 + 1 4 A) −6 B) −7 C) −5 D) −4 13. (03-11-64) Tengsizlikning eng katta butun yechi- mini ko’rsating. 2x − 7 6 + 7x − 2 3 < 3 − 1 − x 2 A) 2 B) −1 C) 1 D) 0 5.2 Chiziqli tengsizliklar sistemasi 1. (97-7-25) Tengsizliklar sistemasini yeching. ½ 3x + 7 ≥ 5(x + 1) + 6 (x − 2) 2 − 8 < x(x − 2) + 10 A) (−11; 2] B) [−2; 7) C) (−7; −2] D) [2; 11) Yechish: Qavsrlarni ochamiz: ½ 3x + 7 ≥ 5x + 5 + 6 x 2 − 4x + 4 − 8 < x 2 − 2x + 10 Endi x qatnashgan hadlarni tengsizlikning chap qismiga, sonlarni tengsizlikning o’ng qismiga o’tka- zib quyidagi sistemani olamiz: ½ −2x ≥ 4 −2x < 14 ⇐⇒ ½ x ≤ −2 x > −7 ⇐⇒ −7 < x ≤ −2. Demak, x ∈ (−7; −2]. Javob: (−7; −2] (C). 2. (97-3-25) Tengsizliklar sistemasini yeching. ½ 2x − 3(x − 5) > 10 − 3x x(x + 2) − 4 ≤ (x − 1) 2 + 7 A) [2; 12, 5) B) [2, 5; ∞) C) [−3; 2) D) (−2, 5; 3] 3. (96-7-25) Tengsizliklar sistemasini yeching. ½ x(x + 1) + 10 > (x + 1) 2 + 3 3x − 4(x − 7) ≥ 16 − 3x A) [−3; 5) B) (2; 4] C) [−6; 6) D) [6, ∞) 4. (97-10-25) Tengsizliklar sistemasini yeching. ½ 4(x − 3) − 3 > 8x + 1 2 + x(x + 3) ≤ (x + 2) 2 + 5 A) (4; 7] B) (−∞; −7) C) (−4; ∞) D) [−7; −4) 5. (01-8-14) Tengsizliklar sistemasini yeching. ( 3x − 2 4 > 1 − 5x 6 3x − 1 ≤ 3 − 2x A) ( 8 19 ; ∞) B) ( 8 19 ; 4 5 ] C) (−∞; 4 5 ] D) x ∈ R 6. (97-6-14) Tengsizliklar ½ 7x + 3 ≤ 9x − 1 20 − 3x ≥ 4x − 15 sistemasi butun yechimlarining o’rta arifmetigini toping. A) 3, 5 B) 7 C) 4 D) 3 Yechish: Sistemada x qatnashgan hadlarni teng- sizlikning chap qismiga, ozod hadlarni tengsiz- likning o’ng qismiga o’tkazib, berilgan sistemaga teng kuchli bo’lgan quyidagi sistemani olamiz: ½ −2x ≤ −4 −7x ≥ −35 ⇐⇒ ½ x ≥ 2 x ≤ 5 ⇐⇒ 2 ≤ x ≤ 5. So’nggi tengsizlikni butun yechimlari 2, 3, 4, 5 lar- dir. Ularning o’rta arifmetigi (2 + 3 + 4 + 5) : 4 = 14 : 4 = 3, 5. Javob: 3, 5 (A). 7. (97-1-14) Tengsizliklar sistemasi butun yechim- larining o’rta arifmetigini toping. ½ 5x − 2 ≥ 2x + 1 2x + 3 ≤ 18 − 3x A) 3 B) 2, 5 C) 2 D) 1, 5 8. (97-11-14) Tengsizliklar sistemasi butun yechim- larining o’rta arifmetigini toping. ½ 2x − 1 ≥ 3x − 4 8x + 7 > 5x + 4 A) 2 B) 2, 5 C) 1, 5 D) 0, 75 9. (02-2-11) Tengsizliklar sistemasi butun yechim- larining o’rta arifmetigini toping. ½ 2x − 10 > 0 27 − x > 0 A) 16 B) 18 C) 17 D) 15 72 10. (96-1-22) Tengsizliklar sistemasi nechta butun yechimga ega? ½ 3 + 4x ≥ 5 2x − 3(x − 1) > −1 A) 5 B) 3 C) 4 D) 2 11. (96-6-16) Tengsizliklar sistemasining eng katta bu- tun yechimini toping. ½ −2x < 22 x + 4 < 8 A) 4 B) 3 C) −11 D) −12 12. (96-9-73) Tengsizliklar sistemasi nechta butun yechimga ega? ½ 3 − 4x > 5 2 + 3(x − 1) ≤ 4x + 3 A) 1 B) 2 C) 4 D) 6 13. (97-2-16) Tengsizliklar sistemasining eng kichik butun yechimimini toping? ½ x + 8 < 12 −3x < 15 A) −5 B) −3 C) −6 D) −4 14. (97-8-16) Tengsizliklar sistemasining barcha bu- tun yechimlari ko’paytmasini toping. ½ −4y < 12 y + 6 < 6 A) 2 B) 6 C) −6 D) −2 15. (97-12-15) Tengsizliklar sistemasining eng katta va eng kichik butun yechimlari yig’indisini to- ping. ½ −2x > −26 x − 3 > 1 A) 17 B) 16 C) 18 D) 19 16. (98-3-15) Tengsizliklar sistemasining butun yechim- lari yig’indisini toping? ½ x + 1 < 2x − 4 3x + 1 < 2x + 10 A) 9 B) 5 C) 20 D) 21 17. (98-1-6) Qo’sh tengsizlikni yeching. −3 < 2 − 5x < 1 A) (−1; 0, 2) B) (−1; −0, 2) C) (−0, 2; 1) D) (0, 2; 1) Yechish: Sistemaning har bir qismidan 2 ni ayirib −5 < −5x < −1 ni olamiz. Bu tengsizlikni −5 ga bo’lamiz (bu holda tengsizliklar qarama- qarshisiga o’zgaradi, chunki −5 manfiy son) va natijada 1 > x > 0, 2 yoki 0, 2 < x < 1 ga ega bo’lamiz. Javob: (0, 2; 1) (D). 18. (98-8-6) Qo’sh tengsizlikni yeching. −4 < 2 − 4x < −2 A) (−1, 5; −1) B) (1; 2) C) (0; 1) D) (1; 1, 5) 19. (99-8-9) 5 < x < 98 tengsizlikni qanoatlantiruv- chi va bo’luvchisi 12 ga teng bo’lgan nechta natu- ral son mavjud? A) 8 B) 10 C) 12 D) 6 20. (99-8-79) Tengsizlik nechta natural yechimga ega? 17, 556 : 5, 7 ≤ y < 31, 465 : 3, 5 A) 1 B) 2 C) 4 D) 5 21. (98-8-1) Tengsizlikning barcha natural yechimla- rini toping. 1256 : 314 < 9x − 32 ≤ 2976 : 96 A) 4; 5; 6 B) 5; 6; 7 C) 6; 7; 8 D) 7; 8 22. (98-1-1) Tengsizlikning barcha natural yechimla- rini toping. 6798 : 103 < 54 + 6x < 9156 : 109 A) 2; 3; 4 B) 4; 5; 6 C) 3; 4 D) 4; 5 23. (99-9-24) Tengsizlikning eng katta butun yechimi, eng kichik butun yechimidan qanchaga katta? ½ 2x − 3 ≤ 17 14 + 3x > −13 A) 17 B) 19 C) 16 D) 18 Yechish: Sistemada ozod hadlarni tengsizlikning o’ng qismiga o’tkazib, berilgan sistemaga teng kuchli bo’lgan quyidagi sistemani olamiz: ½ 2x ≤ 20 3x > −27 ⇐⇒ ½ x ≤ 10 x > −9 ⇐⇒ −9 < x ≤ 10. Bu tengsizlikning eng katta butun yechimi 10, eng kichik butun yechimi −8 dir. Ularning farqi 10 − (−8) = 18. Javob: 18 (D). 24. (98-10-40) Sistemaning eng katta butun va eng kichik butun yechimlari yig’indisini toping. ½ 2x − 3 < 17 4x + 6 > 8 A) 8 B) 11 C) 12 D) 10 25. (98-10-63) Tengsizliklar sistemasining butun yechimlari yig’indisini toping. ½ −x − 5 < −2x − 2 −2x + 2 > 3 − 3x A) 0 B) 1 C) 2 D) 3 26. (98-11-25) Tengsizliklar sistemasining butun yechimlari yig’indisini toping. ½ 0, 4(2x − 3) > x − 2 3x − 7 ≥ x − 6 A) 10 B) 5 C) 6 D) 8 73 5.3 Oraliqlar usuli Kasr ratsional tengsizliklarni yechishning eng qo’lay usullaridan biri – oraliqlar usulidir. Bizdan x 2 + 2x − 3 x 2 + 2x − 8 > 0 (5.3) kasr tengsizlikni yechish talab qilinsin. Ma’lumki, kasr musbat bo’lishi uchun uning maxraji va surati bir xil ishorali bo’lishi kerak. Shunday ekan (5.3) tengsizlik quyidagi ikki tengsizliklar sistemasiga teng kuchlidir: ½ x 2 + 2x − 3 > 0 x 2 + 2x − 8 > 0 yoki ½ x 2 + 2x − 3 < 0 x 2 + 2x − 8 < 0 Bu sistemani yechish yana qismlarga ajratiladi. Agar (x − 1)(x − 3)(x − 5) (x + 2)(x + 4)(x + 6) > 0 ko’rinishdagi tengsizliklarni yechish talab qilinsa, ish yanada qiyinlashadi. Shuning uchun bunday ko’rinish- dagi tengsizliklar odatda oraliqlar usuli bilan yechiladi. Oraliqlar usulining mohiyati quyidagicha: Agar P (x) ko’phad bo’lib, uni P (x) = p(x)(x − x 1 ) r 1 (x − x 2 ) r 2 · · · (x − x n ) r n (5.4) ko’rinishda tasvirlash mumkin bo’lsin. Bunda p(x) ha- qiqiy ildizlarga ega bo’lmagan ko’phad bo’lib, x ning barcha qiymatlarida faqat yo musbat yo manfiy qiy- matlar qabul qiladi. Aniqlik uchun p(x) > 0 va x 1 < x 2 < · · · < x n bo’lsin. Agar x > x n bo’lsa, (5.4) dagi hamma ko’paytuvchilar musbat bo’lib, P (x) > 0 bo’ladi. Agar (x − x n ) chiziqli ko’paytuvchi (5.4) da toq (r n = 2m − 1 toq son) daraja bilan qatnashsa, u holda x n−1 < x < x n bo’lganda (5.4) dagi so’nggi ko’paytuvchi manfiy, boshqa hamma ko’paytuvchilar musbat bo’ladi. Bu holda P (x) < 0 bo’ladi. Bunda P (x) ko’phad x n ildizdan o’tganda ishorasini o’zgartiradi deyiladi. Agar (x − x n ) chiziqli ko’paytuvchi (5.4) da juft (r n = 2m juft son) daraja bilan qatnashsa, u holda x n−1 < x < x n bo’lganda (5.4) dagi bar- cha ko’paytuvchilar musbat bo’lib, P (x) > 0 bo’ladi. Bunda P (x) ko’phad x n ildizdan o’tganda ishorasini o’zgartirmaydi deyiladi. Xuddi shunday muhokama qilish usuli bilan quyidagi xulosaga kelish mumkin: P (x) ko’phad, x o’zgaruvchi x k ildizdan o’tganda (x − x k ) chiziqli ko’paytuvchi toq darajada bo’lsa, ishorasini o’zgartiradi va juft darajada bo’lsa, ishorasini o’zgartirmaydi. Ko’phadning bu xos- sasi yuqori darajali tengsizliklarni yechishda foydalani- ladi. Quyidagi ikki misolni qaraymiz: 1-misol. Tengsizlikni yeching. P (x) = (x 2 +2x+3)(x−1) 2 (x−3) 5 (x−7) 9 ≤ 0. (5.5) Yechish: Bu yerda p(x) = x 2 + 2x + 3 = (x + 1) 2 + 2 > 0. Ko’phad ildizlari 1; 3 va 7 larni sonlar o’qida belgilab chiqamiz. Natijada sonlar o’qi (−∞; 1), (1; 3), (3; 7) va (7; ∞) oraliqlarga ajraydi (5.1-chizmaga qarang). Agar x > 7 bo’lsa, (5.5) dagi barcha ko’paytuvchilar musbat bo’lib P (x) > 0 bo’ladi. Endi 3 ≤ x ≤ 7 bo’lsin, u holda (x−7) 9 ≤ 0 bo’lib, qolgan ko’paytuvchilar musbat bo’ladi va P (x) ≤ 0. Endi 1 < x < 3 bo’lsin, u holda (x − 3) toq daraja bilan qatnashgani uchun P (x) ishorasini o’zgartiradi, ya’ni P (x) > 0 bo’ladi. Agar x < 1 bo’lsa, P (x) ko’phad ishorasini o’zgartirmaydi, chunki x − 1 chiziqli ikki had (5.5) da juft daraja bi- lan qatnashyapti, ya’ni P (x) > 0 bo’ladi. P (1) = 0 bo’lganligi uchun 1 nuqta ham (5.5) tengsizlikning yechimi bo’ladi. Demak, (5.5) tengsizlikning yechimi [3; 7] ∪ {1} to’plamdan iborat. 2-misol. Tengsizlikni yeching. (x − 1) 2 (x − 3) 4 (x − 5) 3 (x + 2) 5 (x + 4) 7 (x + 6) 8 > 0. (4) Yechish: Kasr surati va maxrajidagi ko’phadning ildizlari 1, 3, 5 va −2, −4, −6 larni sonlar o’qida bel- gilab chiqamiz va oraliqlarga + va − ishoralarni qo’yib chiqamiz (5.2-chizmaga qarang). Demak, (4) tengsizlik- ning yechimi (−4; −2)∪(5; ∞) to’plamdan iborat ekan. 1. Oraliqlar usuli. 1. (96-3-25) Tengsizlikni yeching. (x − 2)(x + 3) > 0 A) (−∞; 2) ∪ (3; ∞) B) (−∞; −3) ∪ (2; ∞) C) (−∞; −2) ∪ (3; ∞) D) (−∞; ∞) Yechish: Berilgan ifoda nolga aylanadigan −3, 2 nuqtalarni sonlar o’qida belgilab olamiz, natijada sonlar o’qi (−∞; −3), (−3; 2) va (2; ∞) oraliq- larga bo’linadi (5.3-chizmaga qarang). Agar x > 2 bo’lsa, ifodadagi barcha ko’paytuvchilar mus- bat bo’ladi. Ikkinchi (−3; 2) oraliqda ifoda ishora- sini o’zgartiradi, ya’ni manfiy qiymatlar qabul qiladi. Uchinchi (−∞; −3) oraliqda ifoda yana ishorasini o’zgartiradi, ya’ni ifoda musbat qiy- matlar qabul qiladi. Shunday qilib tengsiz- likning yechimi (−∞; −3) ∪ (2; ∞) to’plamdan iborat bo’ladi. Javob: (−∞; −3) ∪ (2; ∞) (B). 74 2. (96-11-26) Tengsizlikni yeching. (x − 1)(x + 2) > 0 A) (−∞; 1) ∪ (2; ∞) B) (0; ∞) C) (−∞; −2) ∪ (1; ∞) D) (−∞; ∞) 3. (96-12-26) Tengsizlikni yeching. (x + 2)(x − 3) > 0 A) (−∞; ∞) B) (−∞; −3) ∪ (2; ∞) C) (0; ∞) D) (−∞; −2) ∪ (3; ∞) 4. (97-5-23) Tengsizlikni yeching. x − 1 x − 2 ≥ 0 A) [1; 2) B) (−∞; 1) ∪ (2; ∞) C) (1; 2) D) (−∞; 1] ∪ (2; ∞) 5. (97-9-23) Tengsizlikni yeching. x − 2 x − 1 ≤ 0 A) (1; 2] B) [1; 2) C) [1; 2] D) (−∞; 1) 6. (97-9-24) Tengsizlikni yeching. (x + 3)(x − 5) x + 1 ≥ 0 A) (−3; −1] ∪ [5; ∞) B) (−3; −1) ∪ [5; ∞) C) [−3; −1) ∪ [5; ∞) D) [−3; −1) 7. (01-3-35) Tengsizlikni yeching. x + 1 x − 2 ≤ 0 A) (−∞; 1] B) [−1; 2) C) (−1; 2] D) (2; ∞) 8. Tengsizlikni yeching. (x − 2)(x − 4)(x − 5) 2 ≤ 0 A) (−∞; 2] B) [2; 4] ∪ {5} C) [2; 4] D) ∅ 9. (97-12-22) Tengsizlikning eng katta va eng kichik butun yechimlari yig’indisini toping. (x + 4)(3 − x) (x − 2) 2 > 0 A) 1 B) −1 C) −2 D) 2 Yechish: Berilgan kasrning surati nolga aylanadi- gan −4 va 3, maxraji nolga aylanadigan 2 larga mos keluvchi nuqtalarni sonlar o’qida belgilab ola- miz va oraliqlar usulini qo’llaymiz (5.4-chizmaga qarang). Chizmadan ma’lum bo’ldiki, bu tengsi- zlikning yechimi (−4; 2) ∪ (2; 3) to’plamdan ib- orat. Bu to’plamdagi eng katta butun son 1, eng kichik butun son esa −3 dir. Ularning yig’indisi 1 + (−3) = −2. Javob: −2 (C). 10. (96-6-23) Tengsizlikning barcha butun yechimlari yig’indisini toping. (y + 6)(y + 2) < 0 A) 12 B) 20 C) −12 D) −20 11. (99-5-13) Tengsizlikning barcha butun yechimlari yig’indisini toping. (x − 1)(x + 1) 2 (x − 3) 3 (x − 4) 4 ≤ 0 A) 6 B) 7 C) 8 D) 9 12. (00-9-21) Tengsizlikning barcha butun yechimlari yig’indisini toping. (x + 3)(x − 2) 2 (x + 1) 3 (x − 5) 4 ≤ 0 A) 1 B) 2 C) 3 D) 4 13. (01-6-15) Tengsizlikning barcha butun sonlardagi yechimlari yig’indisini toping. x − 4 2x + 6 ≤ 0 A) 7 B) 6 C) 8 D) 5 14. (02-4-12) Tengsizlikni qanoatlantiruvchi eng katta butun sonni toping. x + 5 (x + 6) 2 < 0 A) 6 B) −6 C) 5 D) −7 2. Oraliqlar usuliga keltirib yeching. 15. (98-6-23) Tengsizlikni yeching. x 2 − 2x + 3 x − 1 ≥ 0 A) (1; ∞) B) [1; ∞) C) (−∞; 1) D) (−∞; 1] Yechish: Berilgan kasrning surati x 2 − 2x + 3 = (x − 1) 2 + 2 > 0 doim musbat. Shuning uchun uning maxraji ham musbat bo’lishi kerak, ya’ni x − 1 > 0 ⇐⇒ x > 1. Javob: (1; ∞) (A). 16. (99-3-13) Tengsizlikni yeching. x + 2 − x 2 x 3 + 1 ≥ 0 A) (−∞; 2] B) [2; ∞) C) (−∞; −1) ∪ (−1; 2] D) (−1; 2) |
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