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x + 3y = 5
7x − 18y = 1 tenglamalar sistemasi yechimga ega emas? A) 1 B) 2 C) 4 D) ∅ 15. (97-4-23) Agar ½ x + 2y = 2 2x + y = k bo’lsa, k ning qanday qiymatida x+y = 2 tenglik o’rinli bo’ladi? A) 2 B) 4 C) 1 D) 5 Yechish: Sistema tenglamalarini qo’shamiz: 3x + 3y = 2 + k ⇐⇒ 3(x + y) = 2 + k. Bu yerdan hamda x + y = 2 ekanligini hisobga olsak, 3 · 2 = 2 + k ni olamiz. Demak, k = 4 ekan. Javob: 4 (B). 16. (97-9-83) Agar ½ x + 3y = 6 2x + ky = 8 bo’lsa, k ning qanday qiymatida x+y = 2 tenglik o’rinli bo’ladi? A) 0 B) 1 C) 2 D) 4 4.5.3 Ikkinchi darajali tenglamalar sistemasi 1. (97-8-20) Tenglamalar sistemasini yeching. ½ y + 4 = 2 x 2 y = −2 A) (1; −2) B) (−1; −2) C) (1; 2) D) (−1; −2) va (1; −2) Yechish: Sistemaning 1-tenglamasidan y = −2 ekanligini olamiz va uni 2-tenglamaga qo’yamiz x 2 · (−2) = −2 ⇐⇒ x 2 = 1. Bu chala kvadrat tenglamaning yechimlari x 1 = −1, x 2 = 1 lardir. Shunday qilib, sistema (x 1 ; y) = (−1; −2) va (x 2 ; y) = (1; −2) yechimlarga ega. Javob: (−1; −2) va (1; −2) (D). 2. (96-9-70) Sistemaning yechimini toping. ½ x 2 + y 2 − 2xy = 16 x + y = −2 A) (1; −3) B) (−3; 1) C) (0; −2) D) (1; −3) va (−3; 1) 3. (96-10-20) Sistemaning yechimini toping. ½ x − y = 4 x 2 + y 2 + 2xy = 4 A) (3; 1) B) (3; −1) C) (3; −1) va (1; −3) D) (2; −2) 4. (97-12-19) Tenglamalar sistemasini yeching. ½ x 2 − y 2 + 2x + 4 = 0 x − y = 0 A) (2; 2) B) (−2; −2) C) (−1; −1) D) (1; 1) 5. (97-2-20) Tenglamalar sistemasini yeching. ½ x + 2 = 0 xy 2 = −8 A) (−2; −2) B) (−2; 2) C) (−2; 2) va (−2; −2) D) (2; 2) 6. (96-1-19) Sistemaning yechimini toping. ½ x 2 + y 2 − 2xy = 1 x + y = 3 A) (2; 1) B) (1; 2) C) (1, 5; 1, 5) D) (2; 1) va (1; 2) 67 7. (96-6-20) Tenglamalar sistemasini yeching. ½ x 2 − y 2 − 3x = 12 x − y = 0 A) (−4; 4) B) (4; −4) C) (4; 4) D) (−4; −4) 8. (01-3-34) Tenglamalar sistemasini yeching. ½ x + 3 = 0 xy 2 = −12 A) (−3; 2) B) (−3; −2) C) (−3; −2), (−3; 2) D) ∅ 9. (96-7-23) Tenglamalar sistemasi nechta yechimga ega? ½ x 2 + y 2 = 9 y − x = −3 A) 1 B) 2 C) 3 D) 4 Yechish: Sistemaning 2-tenglamasidan y = x−3 ni topamiz va uni 1-tenglamaga qo’yamiz x 2 + (x − 3) 2 = 9 ⇐⇒ 2x 2 − 6x = 0. Bu chala kvadrat tenglamaning yechimlari x 1 = 0, x 2 = 3 lardir. Bu qiymatlarni y = x − 3 ga qo’yib y 1 = −3, y 2 = 0 ni olamiz. Demak, sis- tema ikkita (x 1 , y 1 ) = (0, −3) va (x 2 , y 2 ) = (3, 0) yechimlarga ega. Javob: 2 (B). 10. (97-10-23) Tenglamalar sistemasi nechta yechimga ega? ½ x 2 + y 2 = 4 x − y = −2 A) 4 B) 3 C) 2 D) 1 11. (97-3-23) Tenglamalar sistemasi nechta yechimga ega? ½ x 2 + y 2 = 25 x − y = 5 A) 4 B) 3 C) 2 D) 1 12. (98-12-19) Agar a − b = 12 va −ab + a 2 = 144 bo’lsa, a ning qiymati qanchaga teng bo’ladi? A) 12 B) −12 C) 36 D) 6 Yechish: 2-tenglikni a(a − b) = 144 shaklda yozamiz va a − b = 12 ekanligidan foydalansak 12a = 144 ni olamiz. Bu tenglikning ikkala qis- mini 12 bo’lib a = 12 ni olamiz. Javob: 12 (A). 13. (98-11-60) Agar x 2 + y 2 = 281 va x − y = 11 bo’lsa, xy qanchaga teng bo’ladi? A) 80 B) 160 C) −80 D) 40 14. (02-12-30) Agar x 2 −4xy+y 2 = 4−2xy va x+y = 12 bo’lsa, xy ning qiymatini toping. A) 32 B) 35 C) 30 D) 34 15. (97-9-67) Agar ab = 9 va 3b = 8c (b 6= 0) bo’lsa, ac ni hisoblang. A) 3 1 3 B) 3 5 8 C) 3 4 9 D) 3 3 8 16. (01-3-33) Sistemadan 3xy ni toping. ½ x 2 + y 2 − xy = 1 x + y = −2 A) 1 B) −1 C) 3 D) −3 Yechish: Berilgan sistema ½ x 2 + y 2 = 1 + xy x + y = −2 sistemaga teng kuchli. Bu sistemaning 2-tenglama- sini (har ikkala qismini) kvadratga ko’taramiz (x + y) 2 = (−2) 2 ⇐⇒ x 2 + y 2 = 4 − 2xy. Endi sistemaning 1-tenglamasidan foydalansak 1+ xy = 4 − 2xy ni olamiz. Bu tenglikdan 3xy = 3 ni olamiz. Javob: 3 (C). 17. (01-3-32) Sistemadan x ni toping. ½ x + y = 6 x 2 − y 2 = 12 A) 4 B) 2 C) 1 D) 3 18. (96-3-75) Sistemadan x ni toping. ½ x + y = 3 x 2 − y 2 = 6 A) 1, 5 B) 2, 5 C) 3 D) 1 19. (96-12-73) Sistemani yeching va x · y ning qiyma- tini toping? ½ x 2 + y 2 = 3 x − y = 1 A) 2 B) 3 C) 1, 5 D) 1 20. (98-12-64) Agar ½ x + y = 3 x · y = 1 bo’lsa, x 5 · y + x · y 5 ni hisoblang. A) 47 B) 29 C) 51 D) 24 Yechish: x + y = 3, xy = 1 tengliklardan x 2 + y 2 = (x + y | {z } 3 ) 2 − 2 xy |{z} 1 = 9 − 2 = 7 ekani kelib chiqadi. Shuning uchun x 5 y + xy 5 = xy(x 4 + y 4 ) = x 4 + y 4 = = (x 2 + y 2 ) 2 − 2(xy) 2 = 7 2 − 2 = 47 Javob: 47 (A). 21. (01-4-23) Agar ½ x 2 − y 2 = 6 x + y = 1 bo’lsa, x − y ning qiymatini toping. A) 1 B) −1 C) 6 D) −6 22. (03-12-3) Sistemadan ab ni toping ½ b + a = 18 a 2 + b 2 = 170 A) 45 B) 72 C) 77 D) 80 68 4.5.4 Ikkinchi va undan yuqori darajali tengla- malar sistemasi 1. (98-10-17) Tenglamalar sistemasini yeching. ½ x 2 − 1 = 0 xy 2 = −4 A) (−1; 2) B) (2; −1) C) (2; 1) D) (−1; −2) va (−1; 2) Yechish: Sistemaning 1-tenglamasidan x 1 = −1, x 2 = 1 ni olamiz. Ikki holni alohida qaraymiz: 1) x 1 y 2 = −4 ⇐⇒ −y 2 = −4 ⇐⇒ y 2 = 4. Bu tenglama y 1 = −2 va y 2 = 2 yechimlarga ega. Demak, (x 1 ; y 1 ) = (−1; −2) va (x 1 ; y 2 ) = (−1; 2) juftliklar berilgan sistemaning yechimlari bo’ladi. 2) x 2 y 2 = −4 ⇐⇒ y 2 = −4. Bu tenglama haqiqiy ildizlarga ega emas. Shun- day qilib sistemaning yechimlari (−1; −2) va (−1; 2) juftliklar ekan. Javob: (−1; −2) va (−1; 2) (D). 2. (02-9-11) Sistema nechta yechimga ega? ( y = 4 x y = −x 2 + 6x − 5 A) ∅ B) 1 C) 2 D) 3 3. (03-9-6) Sistemaning yechimlaridan iborat bar- cha x va y larning yig’indisini toping. ½ x 3 + y 3 = 35 x 2 y + xy 2 = 30 A) 0 B) 2 C) 6 D) 10 4. (02-11-27) x ning ½ x 5 · y 7 = 32 x 7 · y 5 = 128 tenglamalar sistemasining yechimidan iborat bar- cha qiymatlari yig’indisini toping. A) 0 B) 4 C) 8 D) 12 5. (97-4-25) Agar ½ x 3 − y 3 − 3x 2 y = 5 xy 2 = 1 bo’lsa, x − y 2 ni hisoblang. A) 2 B) 1 C) 3 D) 4, 5 Yechish: Berilgan tengliklardan foydalanib (x − y) 3 = x 3 − y 3 − 3x 2 y | {z } 5 +3 xy 2 |{z} 1 = 5 + 3 · 1 = 8 ni hosil qilamiz. U holda x − y = 2 va x − y 2 = 2 2 = 1 bo’ladi. Javob: 1 (B). 6. (98-2-16) Agar m 2 − mn = 48 va n 2 − mn = 52 bo’lsa, m − n nechaga teng? A) 10 B) 8 C) ±10 D) ±8 7. (98-5-22) Agar ½ x 2 − 2xy + y 2 = 9 xy = 10 bo’lsa, |x + y| ni hisoblang. A) 7 B) 6 C) 5 D) 8 8. (98-10-65) (x + y) 2 ni toping. ½ x 2 + y 2 = 10 xy = 3 A) 13 B) 7 C) 16 D) 19 9. (99-6-37) Agar ab = 18, bc = 25 va ac = 8 bo’lsa, √ abc nimaga teng. A) 2 √ 15 B) 15 √ 2 C) 6 √ 5 D) 8 √ 3 10. (99-10-11) Agar x 2 · y = 50, x · y 2 = 20 bo’lsa, xy ning qiymatini hisoblang. A) 8 B) 10 C) 6 D) 12 11. (00-1-23) Agar a − b = 1 va (a 2 − b 2 ) · (a − b) = 9 bo’lsa, ab ning qiymatini toping. A) 19 B) 22 C) 21 D) 20 12. (97-8-11) Agar (x − 4) 2 + (x − y 2 ) 2 = 0 bo’lsa, x + 2y nechaga teng? A) 0 B) 4 C) 6 D) 0 yoki 8 13. (00-8-14) Agar xy = 6 yz = 12 zx = 8 bo’lsa, x + y + z ning qiymatini toping. A) −9 yoki 9 B) 18 C) 0 D) 36 Yechish: Tengliklarni ko’paytirib, x 2 y 2 z 2 = 576 ekanini hosil qilamiz. Ikkita hol bo’lishi mumkin: 1) xyz = 24. Bu tenglikni sistemaning birinchi, ikkinchi va uchinchi tengliklariga ketma-ket bo’lib, z = 4; x = 2; y = 3 ni hosil qilamiz. Bu holda x + y + z = 2 + 3 + 4 = 9. 2) xyz = −24. Bu tenglikni ham sistemaning birinchi, ikkinchi va uchinchi tengliklariga bo’lib, z = −4; x = −2; y = −3 ni hosil qilamiz. Bu holda x + y + z = −9. Javob: −9 yoki 9 (A). 14. (03-8-40) Sistemadan x ni toping. xy x + y = 10 7 yz y + z = 40 13 zx x + z = 5 8 A) 80 79 B) 5 7 C) 7 13 D) 79 80 69 15. (98-6-2) Agar xy = 6, yz = 2 va xz = 3 (x > 0) bo’lsa, xyz ni toping. A) −6 B) 6 C) 5 D) 12 16. (98-11-52) Agar xy = 4, yz = 7 va xz = 28 (y > 0) bo’lsa, xyz ni toping. A) −28 B) 28 C) 27 D) 56 17. (98-6-11) Agar m va n natural sonlar √ 2(n − 5) + n 2 − 6mn + 5m = 0 tenglikni qanoatlantirsa. n − m ni toping. A) 2 B) 5 C) 6 D) 4 Yechish: Agar √ 2(n − 5) noldan farqli bo’lsa, u holda √ 2(n−5)+n 2 −6mn+5m ifoda ham noldan farqli bo’ladi. Chunki irratsional son plyus bu- tun son bu irratsionaldir (nol irratsional emas). Shuning uchun √ 2(n − 5) + n 2 − 6mn + 5m = 0 tenglikdan √ 2(n − 5) = 0 va n 2 − 6mn + 5m = 0 tengliklar kelib chiqadi. 1-tenglikdan n = 5 ni olamiz. Uni 2-tenglikka qo’yib 25−30m+5m = 0 chiziqli tenglamaga ega bo’lamiz. Uning yechim m = 1 dir. Demak, n − m = 5 − 1 = 4. Javob: 4 (D). 18. (01-10-8) Nechta butun x va y sonlar jufti x 2 − y 2 = 31 tenglikni qanoatlantiradi? A) ∅ B) 1 C) 2 D) 4 19. (99-10-22) Agar x 2 ·y+x·y 2 = 48 va x 2 ·y−x·y 2 = 16 bo’lsa, x y ning qiymatini hisoblang. A) 1 4 B) −2 C) 2 D) − 1 2 20. (02-6-32) Sistemadan x · y ni toping ½ x 3 + y 3 = 35 x + y = 5 A) 3 B) 4 C) 5 D) 6 21. (02-7-54) Agar 8a 3 − b 3 = 37 va ab 2 − 2a 2 b = −6 bo’lsa, 2a − b ning qiymatini toping. A) 1 B) −1 C) 2 D) −2 22. (02-8-11) ½ xy + x + y = 11 x 2 y + y 2 x = 30 tenglamalar sistemasi uchun x + y ning eng katta qiymatini toping. A) 6 B) 5 C) 7 D) 4 Yechish: Sistemaning 1-tenglamasidan xy = 11− (x + y) ekanligini topamiz. Agar x + y ni t desak, u holda 2-tenglama xy(x + y) = 30 ⇐⇒ (11 − t)t = 30 ko’rinishga keladi. Bu kvadrat tenglamaning ildiz- lari t 1 = 5, t 2 = 6 lardir. x + y = t bo’lganligi uchun, uning katta qiymati 6. Javob: 6 (A). 23. (02-12-26) Agar 1 n + 1 m = 1 7 va m + n = −4 bo’lsa, mn ning qiymatini toping. A) 20, 5 B) −20, 5 C) 21 D) −28 24. (02-12-29) Agar x 3 + 3xy 2 = 185 va y 3 + 3x 2 y = 158 bo’lsa, x − y ning qiymatini toping. A) 4 B) 3, 5 C) 2 D) 3 25. (03-11-3) Agar 1 √ x + 1 √ y = 4 3 xy = 9 bo’lsa, x + y ning qiymatini toping. A) 10 B) 9 C) 8 D) 12 Yechish: Sistemaning 2-tenglamasi xy = 9 dan foydalanib uning 1-tenglamasini √ x + √ y √ 9 = 4 3 ⇐⇒ √ x + √ y = 4 ko’rinishda yozib olamiz. Bu tenglikning ikkala qismini kvadratga ko’tarib va yana bir marta xy = 9 dan foydalanib x + y + 2 √ xy = 16 ⇐⇒ x + y = 10 ni olamiz. Javob: 10 (A). 26. (03-1-65) Agar ½ x 2 y + xy 2 = 120 x 2 y − xy 2 = 30 bo’lsa, x 2 − y 2 ning qiymatini hisoblang. A) 16 B) 20 C) 25 D) 34 27. (03-11-65) Agar x + 3y + 1 y − y − x + 3 2(x − 2) = 2 y − x = 1 bo’lsa, x · y ning qiymatini toping. A) 15 B) −6 C) −8 D) 12 28. (07-112-29) ½ x 3 + y 3 = 126 x 2 y + xy 2 = 30. tenglamalar sis- temasining haqiqiy yechimlaridan iborat barcha x va y larning yig’indisini toping. A) 2 B) 12 C) 10 D) 6 29. (98-6-10) Agar x 2 + y 2 = 225 va x 2 − y 2 = 63 bo’lsa, |x| − |y| ni toping. A) 3 B) 4 C) 5 D) 6 30. (98-9-16) Agar p 2 + pq = 96 va q 2 + pq = 48 bo’lsa, p + q ning qiymati qanchaga teng bo’ladi? A) 12 B) 14 C) ±12 √ 2 D) ±12 31. (99-1-19) Tenglamalar sistemasini yeching. ½ y − x 3 = 0 y = 16x A) (0; 0), (4; 64) va (−4; −64) B) (0; 0), (8; 2) va (27; 3) C) (0; 0), (2; 8) va (64; 4) D) ∅ 70 5 - bob. Tengsizliklar Agar ikkita ifoda > yoki < (≥ yoki ≤) belgisi bilan bog’langan bo’lsa, ular tengsizlik hosil qiladi deyiladi. Masalan, 3x + 5 > 0; 13x + 1 < 5; x 2 + 7 < 0. Agar tengsizlik > yoki < belgisidan iborat bo’lsa, uni qat’iy tengsizlik deymiz. Masalan, x − 5 < 7; x 2 − 3 > 0. Agar tengsizlik ≥ yoki ≤ belgisidan iborat bo’lsa, uni qat’iymas tengsizlik deymiz. Masalan, x − 9 ≥ 7; x 2 − 5 ≤ 0. Tengsizlik belgisining chap tomonidagi ifodaga tengsizlikning chap qismi, o’ng tomonidagi ifodaga teng- sizlikning o’ng qismi deyiladi. Tengsizlikning ikkala qismi ham sonlardan iborat bo’lsa, unga sonli tengsiz- lik deyiladi. a < b, c < d tengsizliklar bir xil ma’noli, a < b, c > d tengsizliklar esa har xil ma’noli tengsiz- liklar deyiladi. Sonli tengsizliklar quyidagi xossalarga ega. 1. Agar a > b bo’lsa, u holda b < a bo’ladi. 2. Agar a > b va b > c bo’lsa, u holda a > c bo’ladi. 3. Agar a > b va c − ixtiyoriy son bo’lsa, u holda a + c > b + c bo’ladi. Xuddi shunday a > b dan a − c > b − c kelib chiqadi. 4. Agar a > b bo’lib, c > 0 ixtiyoriy son bo’lsa, u holda a · c > b · c hamda a : c > b : c bo’ladi. Ya’ni tengsizlikning ikkala qismini musbat songa ko’paytirsak, yoki bo’lsak tengsizlik belgisi saqlanadi. 5. Agar a > b bo’lib, c < 0 ixtiyoriy son bo’lsa, u holda a · c < b · c hamda a : c < b : c bo’ladi. Ya’ni tengsizlikning ikkala qismini manfiy songa ko’paytirsak, yoki bo’lsak tengsizlik belgisi qarama-qarshisiga o’zgaradi. 6. Bir xil ma’noli tengsizliklarni hadlab qo’shish mumkin, ya’ni a < b va c < d bo’lsa, a + c < b + d bo’ladi. 7. Har xil ma’noli tengsizliklarni hadlab ayirish mumkin, ya’ni a > b va c < d bo’lsa, a − c > b−d bo’ladi. Ayirmada kamayuchi tengsiz- likning belgisi saqlanadi. 8. Nomanfiy hadli bir xil ma’noli tengsizlik- larni hadlab ko’paytirish mumkin, ya’ni 0 ≤ a < b va 0 ≤ c < d bo’lsa, a · c < b · d bo’ladi. 9. Nomanfiy hadli tengsizliklarning har ikkala qismini bir xil natural darajaga ko’tarish mumkin, ya’ni 0 ≤ a < b va n ∈ N bo’lsa, a n < b n bo’ladi. Agar bir tengsizlikning har qanday yechimi shu o’zgaruv- chilar qatnashgan ikkinchi tengsizlikning ham yechimi bo’lsa, va aksincha, ikkinchi tengsizlikning ham qanday yechimi birinchi tengsizlikning yechimi bo’lsa, bu teng- sizliklar teng kuchli tengsizliklar deyiladi. Yechimga ega bo’lmagan tengsizliklar ham teng kuchli tengsiz- liklar hisoblanadi. 5.1 Chiziqli tengsizliklar Download 1.09 Mb. Do'stlaringiz bilan baham: |
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