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m = 1 B) m = 0 C) m = −1 D) m = 2
12. (97-10-22) n ning qanday qiymatlarida nx + 5 = n − 2x tenglamaning ildizi mavjud emas? A) 5 B) −2 C) 1 D) −5 13. (99-7-21) a ning (a 2 − 4)x + 5 = 0 tenglama yechimga ega bo’lmaydigan barcha qiy- matlari ko’paytmasini toping. A) 4 B) −4 C) 0 D) 2 14. (99-8-21) Tenglama a ning qanday qiymatida yechimga ega emas? 6x − a − 6 = (a + 2)(x + 2) A) 4 B) 2 C) −2 D) 6 15. (00-3-11) k ning qanday qiymatida k(k + 6)x = k + 7(x + 1) tenglama yechimga ega bo’lmaydi? A) 1 va 7 B) 1 C) 7 D) 1 va −7 16. (02-7-7) a ning qanday qiymatida 3x − a 5 = ax − 4 3 tenglama ildizga ega emas? A) 1, 8 B) 2 C) 2, 2 D) 1 17. (02-11-9) Tenglamani yeching. 2x + 3 2 + 2 − 3x 3 = 2, 1(6) A) ∅ B) 2 C) −2 D) x ∈ R Yechish: Tenglamaning ikkala qismini 6 ga ko’- paytiramiz: 6x + 9 + 4 − 6x = 6 · 2, 1(6). 2, 1(6) davriy kasrni oddiy kasrga aylantirsak, u 2 16 − 1 90 = 2 1 6 = 13 6 ko’rinish oladi. Natijada (6 − 6)x = 13 − 13 tenglikka ega bo’lamiz. Bu tenglik barcha x ∈ R lar uchun o’rinli. Javob: x ∈ R (D). 18. (03-8-11) Tenglamani yeching. 6x + 2 4 + 2x + 3 2 − 2, 5x + 2 = 4 A) ∅ B) x ∈ R C) 10 D) −10 19. Tenglamani yeching. x + 2 3 + 7x − 1 2 = 3, 8(3)x + 0, 1(6) A) ∅ B) x ∈ R C) 10 D) −10 20. (98-1-20) m ning qanday qiymatlarida m(mx − 1) = 9x + 3 tenglama cheksiz ko’p ildizga ega? A) m = 0 B) m = 3 C) m = −3 D) m = −1 Yechish: Qavslarni ochamiz. m 2 x − m = 9x + 3 Uni (m 2 −9)x−(m+3) = 0 ko’rinishga keltiramiz. Bu tenglama cheksiz ko’p yechimga ega bo’lishi uchun (2-qoidaga qarang) ½ m 2 − 9 = 0 m + 3 = 0 bo’lishi kerak. Demak, m = −3. Javob: −3 (C). 21. (96-7-22) a ning qanday qiymatlarida ax − a = x−1 tenglama cheksiz ko’p yechimga ega bo’ladi? A) a = 1 B) a = 2 C) a = −1 D) a ∈ R 54 22. (96-10-21) n ning qanday qiymatlarida nx + 1 = n+x tenglama cheksiz ko’p yechimga ega bo’ladi? A) n = 0 B) n = 1 C) n = 2 D) n 6= 1 23. (97-7-22) m ning qanday qiymatlarida m 2 x−m = x + 1 tenglama ildizlari cheksiz ko’p bo’ladi? A) m = 1 B) m = 0 C) m = −1 D) m = ±1 24. (98-12-28) Tenglama a ning qanday qiymatlarida cheksiz ko’p yechimga ega? 10(ax − 1) = 2a − 5x − 9 A) − 1 2 B) 2 C) 1 2 D) −2 25. (01-1-10) a ning qanday qiymatida (a 2 + 2)x = a(x − a) + 2 tenglamaning ildizlari cheksiz ko’p bo’ladi? A) − √ 2 B) √ 2 C) √ 2; − √ 2 D) ∅ 26. (03-7-44) Tenglamani yeching. 3 + 25x 3x + 7 = 5 A) −3, 2 B) 1, 5 C) −1 1 5 D) 3, 2 27. (03-7-48) Tenglamani yeching. ³ 1, 7 : ³ 1 2 3 · x − 3, 75 ´´ : 8 25 = 1 5 12 A) 5, 2 B) 5 3 4 C) 4 D) 4, 5 28. (99-8-11) Tenglamani yeching. (x − 12) : 3 8 0, 3 · 3 1 3 + 7 = 1 A) 25 B) 14 C) 15 D) 16 29. (01-8-4) Tenglamani yeching. ³ 4 3 8 x + 5 1 16 ´ · 4 15 = 5 12 x + 2 2 5 A) 1 15 B) 1 2 5 C) 3 185 D) 2 1 5 30. (02-7-43) 986 2 − 319 2 = 2001 · n bo’lsa, n ning qiymatini toping. A) 435 B) 443 C) 515 D) 475 31. (03-11-57) Tenglamani yeching. 12 ³ 1 3 4 x + 5 8 ´ = −6 1 2 A) − 1 3 B) − 2 3 C) 2 3 D) − 13 21 32. (02-7-6) m ning qanday qiymatida 6x − m 2 = 7mx + 1 3 tenglamaning ildizi nolga teng bo’ladi? A) − 2 3 B) 4 5 C) − 3 2 D) 1 2 4.2.1 Proporsiya Proporsiya xossalari 1.2.5-bandda keltirilgan. 1. (96-9-75) Proporsiyaning noma’lum hadini toping. 3 3 5 : 2 7 10 = 3 3 4 : x A) 2 13 16 B) 2 3 10 C) 3 1 3 D) 1 15 16 Yechish: Proporsiya o’rta hadlarining ko’paytmasi uning chetki hadlari ko’paytmasiga tengligidan 2 7 10 · 3 3 4 = x · 3 3 5 ⇐⇒ 27 10 · 15 4 : 18 5 = x ni olamiz. Ko’paytirish va bo’lishni bajarib x = 2 13 16 ekanligini olamiz. Javob: 2 13 16 (A). 2. (00-5-10) Tenglamani yeching. 1 1 12 x : 2 1 12 = 2 3 5 A) 5 B) 3 C) 1 5 12 D) 4 3. (98-12-12) Tenglamani yeching. (12, 5 − x) : 5 = (3, 6 + x) : 6 A) 5 2 11 B) 5 3 11 C) 5 4 11 D) 5 1 11 4. (96-7-12) Proporsiyaning noma’lum hadini toping. 6, 9 : 4, 6 = x : 5, 4 A) 7, 1 B) 7, 7 C) 8, 1 D) 8, 4 5. (97-3-12) Proporsiyaning noma’lum hadini toping. 3, 5 : x = 0, 8 : 2, 4 A) 10, 5 B) 9, 2 C) 13, 5 D) 7, 8 6. (97-7-12) Proporsiyaning noma’lum hadini toping. 5, 4 : 2, 4 = x : 1, 6 A) 3, 6 B) 4 C) 2, 8 D) 4, 6 7. (97-10-12) Proporsiyaning noma’lum hadini to- ping. 0, 25 : 1, 4 = 0, 75 : x A) 3, 6 B) 2, 4 C) 4, 2 D) 5, 2 8. (98-7-13) Tenglamani yeching. ³ 1 3 + x ´ : 7 = ³ 3 4 + x ´ : 9 A) 1 3 8 B) 1 1 8 C) 1 5 8 D) 1 7 8 9. (03-11-55) Proporsiyaning noma’lum hadini to- ping. 12 1 2 : 2 1 2 = 16 2 3 : y A) 3 1 3 B) 3 2 3 C) 3 1 6 D) 3 5 6 55 4.3 Kvadrat tenglamalar ax 2 + bx + c = 0 (4.5) ko’rinishdagi tenglamaga yoki ayniy almashtirishlar- dan keyin (4.5) ko’rinishga keltirish mumkin bo’lgan tenglamaga bir noma’lumli ikkinchi darajali tenglama yoki kvadrat tenglama deyiladi. Bunda x − noma’lum, a (a 6= 0), b va c − lar ixtiyoriy sonlar. (4.5) tenglam- aga kvadrat tenglamaning normal shakli deyiladi. a ni birinchi koeffitsiyent, b ni ikkinchi koeffitsiyent, c ni esa uchinchi koeffitsiyent yoki ozod had deyiladi. Kvadrat tenglamaning ildizlari quyidagi formula orqali topiladi x 1 = −b − √ b 2 − 4ac 2a , x 2 = −b + √ b 2 − 4ac 2a (4.6) Bunda b 2 − 4ac ga kvadrat tenglamaning diskriminanti deyiladi va u D harfi bilan belgilanadi: D = b 2 − 4ac. Agar (4.5) tenglamaning birinchi koeffitsiyenti, ya’ni a = 1 bo’lsa bunday tenglama keltirilgan kvadrat teng- lama deyiladi. Keltirilgan kvadrat tenglama odatda x 2 + px + q = 0 (4.7) ko’rinishda yoziladi. Agar (4.5) tenglamada b = 0 yoki c = 0 yoki b = c = 0 bo’lsa, ular chala kvadrat tenglamalar deyiladi. Masalan, ax 2 + bx = 0 chala kvadrat tenglamaning ildizlari x 1 = 0, x 2 = −b/a lardir. ax 2 + c = 0 chala kvadrat tenglama ac < 0 shartda yechimga ega va un- ing ildizlari x 1,2 = ± q −c a , ko’rinishga ega. ax 2 = 0 chala kvadrat tenglamaning ildizi esa x 1 = x 2 = 0 dir. 1. Viyet teoremasi. Agar x 1 , x 2 sonlar (4.7) tenglamaning ildizlari bo’lsa, u holda ½ x 1 + x 2 = −p x 1 · x 2 = q tengliklar o’rinli. 2. Agar D > 0 bo’lsa, u holda (4.5) kvadrat tenglama ikkita har xil haqiqiy ildizga ega. 3. Agar D = 0 bo’lsa, u holda (4.5) kvadrat tenglama bitta haqiqiy ildizga ega. 4. Agar D < 0 bo’lsa, u holda (4.5) kvadrat tenglama haqiqiy ildizlarga ega emas. 5. ax 2 + bx + c kvadrat uchhad ax 2 + bx + c = a(x − x 1 )(x − x 2 ) ko’paytuvchilarga ajraydi. Bunda x 1 , x 2 sonlar (4.5) kvadrat tenglamaning ildizlari. Keltirilgan (4.7) kvadrat tenglamaning ildizlari quyidagi xossalarga ega. 6. x 2 1 + x 2 2 = (x 1 + x 2 ) 2 − 2x 1 x 2 = p 2 − 2q. 7. (x 1 − x 2 ) 2 = (x 1 + x 2 ) 2 − 4x 1 x 2 = p 2 − 4q. 8. x 3 1 + x 3 2 = 3pq − p 3 . 9. x 4 1 + x 4 2 = p 4 − 4p 2 q + 2q 2 . 1. Kvadrat tenglamani yeching. x 2 + 5x − 6 = 0 A) −6; 1 B) −1; 6 C) 1; 6 D) 2; 3 Yechish: Berilgan kvadrat tenglamada a = 1, b = 5, c = −6. Endi kvadrat tenglama yechimga egami yoki yo’qmi shuni aniqlaymiz. Buning uchun un- ing diskriminantini hisoblaymiz: D = 5 2 − 4 · 1 · (−6) = 25+24 = 49. Kvadrat tenglama diskrimi- nanti musbat, shuning uchun u ikkita ildizga ega. Ularni (4.6) formula yordamida hisoblaymiz: x 1 = −5 − √ 49 2 · 1 = −5 − 7 2 = −6, x 2 = −5 + √ 49 2 · 1 = −5 + 7 2 = 1. Javob: −6; 1 (A). 2. Kvadrat tenglamani yeching. 2x 2 + 3x − 14 = 0 A) −7; 2 B) −2; 3 1 2 C) −3 1 2 ; 2 D) 2; 3 1 2 3. Kvadrat tenglamani yeching. 4x 2 + 12x + 9 = 0 A) −7; 2 B) −1, 5 C) −3 D) ∅ 4. Kvadrat tenglamani yeching. 9x 2 + 6x + 3 = 0 A) −1; 1 3 B) −2; 1 6 C) −3 D) ∅ 5. (00-8-64) Tenglamani yeching. 1998x 2 − 2000x + 2 = 0 A) 1; 2 1998 B) −1; 2 1998 C) 1; − 2 1998 D) −1; − 2 1998 6. Tenglamani yeching. x 2 − 97x + 2010 = 0 A) 30; 67 B) −30; −67 C) 15; 134 D) 2; 1005 7. Chala kvadrat tenglamani yeching. 2x 2 − 6x = 0 A) 0; 3 B) −2; 6 C) 3 D) ∅ Yechish: Berilgan kvadrat tenglamada a = 2, b = −6, c = 0. Bu tenglama diskriminantini D = (−6) 2 −4·2·0 = 36 > 0. Kvadrat tenglama ikkita ildizga ega. (4.6) formuladan x 1 = 0, x 2 = 3. Javob: 0; 3 (A). 56 8. Chala kvadrat tenglamani yeching. x 2 − 7x = 0 A) 0; 3 B) 7 C) 0; 7 D) 0 9. Chala kvadrat tenglamani yeching. 9x 2 − 1 = 0 A) 0; 1 3 B) − 1 3 ; 1 3 C) 0; − 1 3 D) ∅ 10. Chala kvadrat tenglamani yeching. 3x 2 + 1 3 = 0 A) 0; 1 3 B) − 1 3 ; 1 C) 0; − 1 3 D) ∅ 11. (96-1-18) Tenglamaning nechta ildizi bor? 3 − x = − 4 x A) 1 B) 2 C) 3 D) ildizi yo’q Yechish: x 6= 0 deb, tenglikning ikkala qismini x ga ko’paytiramiz va 3x−x 2 +4 = 0 ni olamiz. Bu kvadrat tenglamaning diskriminanti D = 3 2 − 4 · (−1) · 4 = 25 > 0. Shuning uchun u ikkita ildizga ega. Javob: 2 (B). 12. (96-9-69) Tenglamaning nechta ildizi bor? 2 x = x + 2 A) 3 B) 2 C) 1 D) ildizi yo’q 13. Tenglamaning nechta ildizi bor? x 2 − 6x + 9 = 0 A) 1 B) 2 C) 3 D) ildizi yo’q 14. Tenglamaning nechta haqiqiy ildizi bor? 4x 2 + 8x + 7 = 0 A) 1 B) 2 C) 3 D) ildizi yo’q 15. (96-7-13) Agar (3x − 1) · (x − 2) = 0 bo’lsa, 3x − 1 qanday qiymatlarni qabul qilishi mumkin? A) faqat 1 3 B) faqat 0 C) 1 3 yoki 0 D) 0 yoki 5 Yechish: (3x−1)·(x−2) = 0 dan x 1 = 1 3 , x 2 = 2 ekanligi kelib chiqadi. Bu qiymatlarni 3x − 1 ga qo’yib 3x 1 − 1 = 0 va 3x 2 − 1 = 5 ni olamiz. Javob: 0 yoki 5 (D). 16. (97-3-13) Agar (2x + 1) · (x − 1, 5) = 0 bo’lsa, 2x + 1 qanday qiymatlar qabul qiladi? A) faqat 0 B) faqat − 1 2 C) 0 yoki − 1 2 D) 4 yoki 0 17. (97-7-13) Agar (x − 5) · ( 1 5 x + 4) = 0 bo’lsa, 1 5 x + 4 qanday qiymatlar qabul qiladi? A) faqat 0 B) faqat −20 C) 0 yoki 5 D) 0 yoki 8 18. (97-10-13) Agar (4x + 1) · (x − 1 4 ) = 0 bo’lsa, 4x + 1 qanday qiymatlar qabul qilishi mumkin? A) faqat − 1 4 B) faqat 1 4 C) faqat 0 D) 0 yoki 2 19. (96-3-18) Kvadrat uchhadni chiziqli ko’paytuvchi- larga ajrating. x 2 − x − 2 A) (x − 1)(x + 2) B) (x − 1)(x − 2) C) (x + 1)(x + 2) D) (x + 1)(x − 2) Yechish: Berilgan kvadrat uchhadni chiziqli ko’- paytuvchilarga ajratish uchun 5-xossadan foydala- namiz. Shu maqsadda kvadrat tenglamaning ildiz- larini topamiz. Uning diskriminanti D = (−1) 2 − 4 · 1 · (−2) = 1 + 8 = 9 ga teng. D > 0 bo’lganligi uchun tenglamaning ikkita ildizi bor. Ularni (4.6) formula yordamida topamiz x 1 = 1 − 3 2 = −1, x 2 = 1 + 3 2 = 2. 5-xossadan x 2 − x − 2 = (x + 1)(x − 2) ni olamiz. Javob: (x + 1)(x − 2) (D). 20. (96-11-19) Kvadrat uchhadni chiziqli ko’paytuvchilarga ajrating. x 2 − 3x + 2 A) (x − 1)(x + 2) B) (x − 2)(x + 1) C) (x − 1)(x − 2) D) (x + 1)(x + 2) 21. (97-2-27) Kasrni qisqartiring. x 2 − 16 x 2 − 5x + 4 A) 4 + x 1 − x B) 4 − x x + 1 C) x + 4 x + 1 D) x + 4 x − 1 22. (97-8-26) Kasrni qisqartiring. y 2 − 3y − 4 y 2 − 1 A) y + 4 y + 1 B) 4 − y y − 1 C) y + 4 y − 1 D) y − 4 y − 1 57 23. (00-8-37) Ko’paytuvchilarga ajrating. 3x 2 − 6xm − 9m 2 A) 3(x + m)(x − 3m) B) (x − 3m) 2 C) 3(x − m)(x + 3m) D) (3x − m) 2 24. (00-3-18) Agar x 2 − 3x − 6 = 0 tenglamaning ildizlari x 1 va x 2 bo’lsa, 1 x 3 1 + 1 x 3 2 ni toping. A) 1 3 B) 0, 5 C) −0, 5 D) −0, 375 Yechish: Berilgan tenglama uchun p = −3, q = −6. Qiymati izlanayotgan ifodani 1 x 3 1 + 1 x 3 2 = x 3 1 + x 3 2 (x 1 x 2 ) 3 shaklda yozib olamiz. Viyet teoremasi va 8-xossaga ko’ra bu kasrning qiymati x 3 1 + x 3 2 (x 1 x 2 ) 3 = 3 · (−3) · (−6) − (−3) 3 (−6) 3 = − 81 216 ga teng. Buni o’nli kasrga aylantirib −0, 375 ni olamiz. Javob: −0, 375 (D). 25. (96-13-18) x 1 va x 2 sonlar x 2 + x − 5 = 0 tenglamaning ildizlari ekanligi ma’lum. x 2 1 + x 2 2 ning qiymatini toping. A) 10 B) 12 C) 11 D) 9 26. (97-4-24) a va b sonlari 3x 2 − 2x − 6 = 0 tenglamaning ildizlari bo’lsa, a 2 +b 2 ni hisoblang. A) 6 B) 8 C) 4 4 9 D) 4 2 9 27. (98-4-25) Agar x 2 + x − 1 = 0 tenglamaning ildizlari x 1 va x 2 bo’lsa, x 3 1 + x 3 2 ning qiymati qanchaga teng bo’ladi? A) 1 B) 3 C) 2 D) −4 28. (98-5-21) Ushbu x 2 + 4x − 5 = 0 tenglamaning ildizlari x 1 va x 2 bo’lsa, x 3 1 · x 3 2 ni hisoblang. A) 124 B) −125 C) 130 D) 5 29. (99-7-23) Agar x 2 + 2x + 1 = 0 tenglamaning ildizlari x 1 va x 2 bo’lsa, x 3 1 − x 3 2 ni hisoblang. A) 1 B) 3 C) 4 D) 0 30. (01-10-2) Agar x 1 va x 2 lar x 2 +x−5 = 0 tenglama- ning ildizlari bo’lsa, x 2 1 x 4 2 + x 4 1 x 2 2 ning qiymatini hisoblang. A) 225 B) 145 C) 125 D) 275 31. (02-11-13) Agar x 1 va x 2 9x 2 + 3x − 1 = 0 tenglamaning ildizlari bo’lsa, 3x 1 x 2 x 1 + x 2 ning qiy- matini toping. A) −1 B) 1 C) 2 D) 1 3 32. (03-1-5) Agar x 1 va x 2 x 2 + 3x − 3 = 0 tenglamaning ildizlari bo’lsa, x 4 1 +x 4 2 ning qiyma- tini hisoblang. A) 207 B) 192 C) 243 D) 168 33. (03-8-19) Tenglamaning katta va kichik ildizlari kublarining ayirmasini toping. Download 1.09 Mb. Do'stlaringiz bilan baham: |
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