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0 tenglama ildizlaridan Q(x) = 0 tenglama ildizlarini chiqarib tashlashdan hosil bo’ladi. Quyidagi misolni qaraymiz: x 2 + 5x + 6 x 2 + x − 2 = 0. Ma’lumki, kasrning qiymati nol bo’lishi uchun uning surati nolga teng bo’lishi kerak. Shuning uchun x 2 + 5x + 6 = 0 tenglamani yechamiz. Bu tenglamaning ildizlari x 1 = −2, x 2 = −3. Topilgan bu qiymatlarni kasr maxrajiga qo’yamiz: (−2) 2 + (−2) − 2 = 4 − 2 − 2 = 0; (−3) 2 + (−3) − 2 = 9 − 3 − 2 = 4 6= 0. Shunday qilib, beril- gan tenglamaning ildizi x = −3 ekan. x = −2 esa tenglamaning ildizi bo’lmaydi. Chunki bu qiymatda kasrning maxraji nolga aylandi. 1. (00-3-26) Tenglamaning haqiqiy ildizlari yig’indi- sini toping. (x 2 + 5x + 4)(x 2 + 5x + 6) = 120 62 A) 3 B) −3 C) 2 D) −5 Yechish: Tenglama y = x 2 + 5x belgilash yor- damida (y + 4)(y + 6) = 120 tenglamaga keladi. Qavslarni ochamiz y 2 + 10y + 24 − 120 = 0 ⇐⇒ y 2 + 10y − 96 = 0. Bu kvadrat tenglamaning yechimlari y 1 = 6, y 2 = −16 lardir. Endi berilgan tenglama ikkita tenglam- aga ajraydi. 1) x 2 +5x = 6, x 2 +5x−6 = 0, x 1 = −6, x 2 = 1, 2) x 2 + 5x + 16 = 0, D = 25 − 64 = −39 < 0. 2) tenglama yechimga ega emas. Demak, beril- gan tenglama x 1 = −6 va x 2 = 1 ildizlarga ega. Ildizlar yig’indisi x 1 + x 2 = −5. Javob: −5 (D). 2. (96-7-15) Tenglama ildizlari yig’indisini toping. x 4 − 13x 2 + 36 = 0 A) 13 B) 5 C) 0 D) 36 3. (97-7-15) Tenglamaning eng katta va eng kichik ildizlari ayirmasini toping. x 4 − 10x 2 + 9 = 0 A) 1 B) 8 C) 2 D) 6 4. (98-4-33) Tenglamaning ildizlari yig’indisini to- ping. 2x 4 − 7x 2 + 2 = 0 A) 7 B) 3, 5 C) 0 D) 2 5. (98-6-20) Tenglama ildizlari ko’paytmasini toping. (x + 1 x ) 2 − 2(x + 1 x ) − 3 = 0 A) 3 B) −1 C) 4 D) 1 6. (98-11-10) Tenglamaning haqiqiy ildizlari ko’payt- masini aniqlang. y 4 − 2y 2 − 8 = 0 A) 4 B) −16 C) 16 D) −4 7. (00-1-16) Ifoda nechta ratsional koeffitsiyentli ko’paytuvchilarga ajraladi? (x 4 + x 2 + 1) · (x 4 + x 2 + 2) − 12 A) 4 B) 2 C) 3 D) 5 8. (98-6-22) Tenglamani yeching. 2x 2 − 5x + 3 (10x − 5)(x − 1) = 0 A) 1 B) 1; 3 2 C) 3 2 D) 5 Yechish: Berilgan tenglamada kasr suratini nolga tenglashtiramiz. 2x 2 −5x+3 = 0 kvadrat tenglama ildizlari x 1 = 1, x 2 = 1, 5 lardir. Berilgan tengla- madagi kasr maxraji x = 1 da nolga aylanadi, x = 1, 5 da esa noldan farqli. Demak, x = 1, 5 tenglamaning ildizi bo’ladi. Javob: 3 2 (C). 9. (98-11-18) Tenglamaning yechimlari quyidagi ora- liqlarning qaysi birida joylashgan? x 2 + 1 x + x x 2 + 1 = −2, 5 A) (−∞; −1) B) [−1; 8) C) [2; 8) D) [3; 8) 10. (98-11-71) Tenglamani yeching. 1 − 1 x−1 1 + 1 x−1 = 0 A) −2 B) 0 C) −1 D) 2 11. (98-12-63) Tenglama ildizlari yig’indisini toping. 2 3 − x + 1 2 = 6 x(3 − x) A) 4 B) 7 C) 3 D) 10 12. (00-5-36) Tenglamaning ildizlari nechta? x 2 − x − 2 x 2 + x = 0 A) 2 B) 4 C) 1 D) 3 13. (01-1-8) Tenglamani yeching. 2 x − 3 = x + 5 x 2 − 9 A) −2 B) 2 C) 1 D) −1 14. (02-3-25) Tenglama ildizlari ko’paytmasini toping. 26 5(x + x −1 ) = 1 A) 1 B) 5 C) 2 D) 2, 4 15. (02-4-4) Tenglamaning ildizlari sonini toping. x 4 − ( √ 5 + √ 3)x 2 + √ 15 = 0 A) 2 B) 4 C) 1 D) 0 16. (02-7-41) Tenglama ildizlari yig’indisini toping. (x + 1)(x + 2)(x + 4)(x + 5) = 40 A) −6 B) 0 C) −5 D) 6 Yechish: Berilgan tenglamaning chap qismidagi ko’paytmani (x + 1)(x + 5) va (x + 2)(x + 4) lar- ning ko’paytmasi shaklida yozamiz va bu qavslar- ni ochamiz. Natijada, berilgan tenglamaga teng kuchli bo’lgan (x 2 + 6x + 5)(x 2 + 6x + 8) − 40 = 0 tenglamaga ega bo’lamiz. Bu tenglamada x 2 + 6x + 5 = y deb belgilash olib y(y + 3) − 40 = 0 ⇐⇒ y 2 + 3y − 40 = 0 63 kvadrat tenglamani hosil qilamiz. Bu tenglama- ning ildizlari y 1 = −8, y 2 = 5 lardir. Topilgan bu qiymatlarni belgilashga qo’yamiz: 1) x 2 + 6x + 5 = −8; 2) x 2 + 6x + 5 = 5. 1-kvadrat tenglamaning diskriminanti D = 6 2 − 4·13 = −14 < 0 manfiy, shuning uchun u haqiqiy ildizlarga ega emas. 2-tenglama x 2 + 6x = 0 ko’rinishdagi chala kvadrat tenglama bo’lib, un- ing ildizlari x 1 = −6, x 2 = 0 lardir. Tenglama ildizlari yig’indisi −6 ekan. Javob: −6 (A). 17. (02-11-20) Tenglama ildizlari yig’indisini toping. 3x 2 + 8x − 3 x + 3 = x 2 − x + 2 A) −8 B) −6 C) −4 D) 4 18. (03-3-28) Tenglama ildizlari ko’paytmasini toping. 3x 2 + 8x − 3 x + 3 = x 2 − x + 2 A) 2 B) −2 C) 6 D) 3 19. (03-6-8) Agar 4x 2 − 4xy + 3y 2 2y 2 + 2xy − 5x 2 = 1 bo’lsa, x + y x − y ning qiymatini toping. A) 2 B) −2 C) 1 2 D) − 1 2 20. (03-7-56) Tenglama ildizlari ayirmasining modu- lini toping. x + 8 3 = x − x − 3 x A) 5, 5 B) 5 C) 3, 5 D) 4 21. (03-8-17) Tenglama ildizlari yig’indisini toping. 3x 2 + 4x − 4 x + 2 = x 2 − 4x + 4 A) 10 B) −5 C) −4 D) 7 22. (03-8-42) Tenglama butun ildizlarining yig’indisini toping. x 2 + 3x + 6 2 − 3x − x 2 = 1 A) −3 B) 1 C) −5 D) 3 23. (03-11-63) Tenglama ildizlari yig’indisini toping. x 3 − 8 x − 2 = 6x + 1 A) 6 B) 4 C) −4 D) 3 24. (03-12-2) Tenglamaning eng kichik va eng katta ildizlari ayirmasini toping. 3x 4 − 5x 2 + 2 = 0 A) 2 B) 2 √ 6 3 C) − 2 √ 6 3 D) −2 4.5 Tenglamalar sistemasi 4.5.1 Chiziqli tenglamalar sistemasi Ikki noma’lumli birinchi darajali (chiziqli) tenglamalar sistemasi deb ½ a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 (4.13) ko’rinishdagi sistemaga aytiladi. Bu yerda a 1 , b 1 , c 1 , a 2 , b 2 , c 2 ma’lum sonlar bo’lib, ular berilgan bo’ladi, x, y lar esa noma’lumlar hisoblanadi. (4.13) sistemaning yechimi deb birinchi va ikkinchi tenglamalarni qanoat- lantiruvchi (x; y) sonlar juftiga aytiladi. Sistemani ye- chish uning hamma yechimlarini topish yoki yechim- lari yo’qligini isbotlashdan iborat. (4.13) sistemani yechishning quyidagi ikki usulini ko’rib o’tamiz. O’rniga qo’yish usuli. Tenglamalar sistemasini o’rni- ga qo’yish usuli bilan yechish uchun tenglamalarning birortasidan noma’lumlardan birini ikkinchisi orqali ifo- dalab, ikkinchi tenglamaga olib borib qo’yiladi. Nati- jada bir noma’lumli chiziqli tenglama hosil bo’ladi. Bu tenglamani yechib, yechimni tenglamalar sistemasining istalgan bir tenglamasiga qo’yib ikkinchi noma’lumning qiymati topiladi. Quyidagi misolni qaraymiz: ½ 2x − 3y = 3 x + 2y = 5. (4.14) Bu sistemaning ikkinchisidan x ni topamiz (ya’ni x = 5−2y) va uni sistemaning birinchi tenglamasiga qo’yamiz: 2(5 − 2y) − 3y = 3 ⇐⇒ 7 = 7y. Bu tenglama yag- ona y = 1 ildizga ega. Bu yechimni sistemaning ikkin- chisiga qo’yib, x+2·1 = 5 dan x = 3 ni olamiz. Demak, (4.14) sistemaning yechimi (3; 1) dan iborat. Qo’shish usuli. Tenglamalar sistemasini qo’shish usuli bilan yechish uchun berilgan sistemaga teng kuchli bo’lgan shunday sistema olamizki, unda y (yoki x) ol- didagi koeffitsiyentlar qarama-qarshi sonlar bo’ladi. Sis- temaning tenglamalari hadma-had qo’shilsa, x ga nis- batan bir noma’lumli chiziqli tenglama hosil bo’ladi. Bu tenglamadan x ni topib, uni tenglamalar sistemasin- ing istalgan bir tenglamasiga qo’yib y noma’lumning qiymati topiladi. Quyidagi misolni qaraymiz: ½ 3x − 4y = 3 x + 2y = 1. (4.15) Bu sistemaning ikkinchi tenglamasini 2 ga kopaytiramiz va tenglamalarni hadma-had qo’shib 5x = 5 tenglamani olamiz. Bu yerdan x = 1 ni topamiz va uni sistemaning ikkinchi tenglamasiga qo’yib 1 + 2y = 1 dan y = 0 ni olamiz. Demak, (4.15) sistemaning yechimi (1; 0) dan iborat. 1. Agar (1) sistemada a 1 a 2 6= b 1 b 2 bo’lsa, u ya- gona yechimga ega. 2. Agar (1) sistemada a 1 a 2 = b 1 b 2 = c 1 c 2 bo’lsa, u cheksiz ko’p yechimlarga ega. 64 3. Agar (1) sistemada a 1 a 2 = b 1 b 2 6= c 1 c 2 bo’lsa, sistema yechimga ega emas. 1. (96-6-17) Agar 3x + y = 45 z + 3y = −15 3z + x = 6 bo’lsa, x + y + z nimaga teng? A) 12 B) 10 C) 15 D) 9 Yechish: Sistema tenglamalarini qo’shamiz: 4x + 4y + 4z = 45 − 15 + 6. Bu yerdan 4(x + y + z) = 36 ⇐⇒ x + y + z = 9 ekanini hosil qilamiz. Javob: 9 (D). 2. (96-1-21) (x; y) sonlar jufti ½ 2x − y = 5 3x + 2y = 4 sistemaning yechimi bo’lsa, x − y ni toping. A) 1 B) −1 C) 3 D) 0 3. (96-3-24) Tenglamalar sistemasini qanoatlantiruvchi sonlar juftligini aniqlang. ½ x + y = 5 x − y = 1 A) (2; 3) B) (−2; 3) C) (3; 2) D) (−2; −3) 4. (96-3-76) x ni toping. ½ 2x − 3y = 3 x + 2y = 5 A) 1 B) 2 C) 3 D) −2 5. (96-9-17) x ni toping. ½ 3x − 4y = 3 x + 2y = 1 A) 1 B) 0 C) −1 D) 2 6. (96-9-72) (x; y) sonlar jufti ½ 3x − 2y = −8 x + 3y = 1 sistemaning yechimi bo’lsa, y − x ni toping. A) 0 B) −1 C) −2, 5 D) 3 7. (96-11-25) Quyidagi juftliklardan qaysi biri tenglamalar sistemasini qanoatlantiradi? ½ x + y = 5 x − y = −1 A) (2; 3) B) (1; 4) C) (4; 1) D) (3; 2) 8. (96-12-74) Sistemani yeching va y ning qiymatini toping. ½ 2x − 3y = 3 x + 2y = 5 A) 2 B) 1 C) 3 D) 1, 5 9. (96-13-17) Sistemadan x ni toping. ½ 3x − 4y = 3 x + 2y = 1 A) −1 B) 3 C) 2 D) 1 10. (97-1-11) (x; y) sonlar jufti tenglamalar sistema- sining yechimi, x · y ni toping. ½ 2x + y − 8 = 0 3x + 4y − 7 = 0 A) −90 B) 12 C) −10 D) 80 Yechish: Sistemaning birinchi tenglamasidan y ni topamiz y = 8 − 2x va uni ikkinchi tenglamaga qo’yamiz: 3x + 4(8 − 2x) = 7 ⇐⇒ 25 = 5x. Bu yerdan x = 5 ni olamiz. x ning bu qiymatini y = 8 − 2x ga qo’yib y = −2 ni olamiz. Ularning ko’paytmasi xy = 5 · (−2) = −10 ekan. Javob: −10 (C). 11. (97-3-21) Agar ½ 5x + 2y = −3 x − 3y = −4 bo’lsa, x 2 − y 2 ning qiymatini toping. A) 2 B) 1 C) 0 D) 2, 5 12. (97-6-11) (x; y) sonlar jufti ½ x + 2y − 3 = 0 2x − 3y + 8 = 0 tenglamalar sistemasining yechimi, x+y ni hisob- lang. A) −1 B) 1 C) 3 D) 4, 5 13. (97-10-21) Agar ½ 3x − 2y = 1 4x − y = −2 bo’lsa, y 2 − x 2 ning qiymatini toping. A) −1 B) −3 C) 3 D) 5 14. (98-3-16) Sistemadan x ni toping. ½ 3x + 4y = 11 5x − 2y = 1 A) 2 B) 3 2 C) 5 2 D) 1 15. (98-10-64) Sistemadan y ni toping. ½ 3x + 4y = 11 5x − 2y = 1 A) 0 B) 1 C) 2 D) −2 65 16. Tenglamalar sistemasini yeching. ½ x + y = 8 x + 2y = 12 A) (4; 4) B) (−4; −4) C) (−4; 4) D) (4; 6) 17. (97-4-7) a = 4b va c + 3b = 0 (b 6= 0) bo’lsa, a c ni toping. A) − 1 3 B) 1 1 3 C) 1 2 3 D) − 4 3 Yechish: Sistemadan a = 4b, c = −3b ekanligi kelib chiqadi. Demak, a c = 4b −3b = − 4 3 . Javob: − 4 3 (D). 18. (97-8-17) Agar 2m+n = 2, 2n+p = 6 va 2p+m = 4 bo’lsa, m + n + p ni toping. A) 6 B) 4 C) 5 D) 3 19. (97-12-16) Agar 2q − 4p = −9, 2t − 4q = −7 va 2p−4t = 2 bo’lsa, p+q +t ning qiymatini toping. A) −7 B) 8 C) 7 D) −8 20. (00-4-39) Agar 3a + 4b = 16 va 2c − b = 1 bo’lsa, 3a + 8c ning qiymatini toping. A) 18 B) 4 C) 20 D) 23 21. (00-1-11) Agar a 2 − 4a + 5 + b 2 − 2b = 0 bo’lsa, (a + b) 3 ning qiymatini toping. A) 26 B) 27 C) 28 D) 25 22. (02-12-2) Agar x + y = 4, y + z = 8 va x + z = 6 bo’lsa, x − y + 2z ning qiymatini hisoblang. A) 8 B) 6 C) 7 D) 10 23. (02-12-19) Nechta natural sonlar jufti x 2 − y 2 = 105 tenglikni qanoatlantiradi? A) 3 B) 4 C) 2 D) 5 4.5.2 Parametrli tenglamalar sistemasi 1. (00-5-27) k ning qanday qiymatida ½ kx + 4y = 4 3x + y = 1 tenglamalar sistemasi yagona yechimga ega bo’ladi? A) k 6= 12 B) k = 9 C) k 6= 19 D) k = 12 Yechish: 1-qoidaga ko’ra, sistema a 1 a 2 6= b 1 b 2 shartda yagona yechimga ega. Bu shart berilgan sistema uchun k 3 6= 4 1 shartdan iborat. Bu yerdan k 6= 12. Javob: k 6= 12 (A). 2. (01-7-19) a ning qanday qiymatida ½ x + y = a xy = 9 tenglamalar sistemasi yagona yechimga ega? A) 3 B) 6 C) −3 D) −6; 6 3. (01-10-28) a ning qanday qiymatida 2x + 3y = 5 x − y = 2 x + 4y = a tenglamalar sistemasi yechimga ega? A) 0 B) 1 C) 2 D) 3 4. (02-1-46) Agar ½ ax + by = 3 bx + ay = 2 tenglamalar sistemasi x = 3, y = 2 yechimga ega bo’lsa, a ning qiymatini toping. A) 4 B) 5 C) 3 D) 1 5. (02-9-15) Agar y − x = 2 va a > 0 bo’lsa, ½ y 2 − x 2 = 8a y + x = a 2 tenglamalar sistemasini yeching. A) (5; 7) B) (7; 9) C) (4; 6) D) (−6; −4) 6. (99-9-16) k ning qanday qiymatida ½ 3x + 6y = k 9x + 18y = k + 1 tenglamalar sistemasi cheksiz ko’p yechimga ega? A) 1 3 B) 1 C) 1 2 D) 2 3 Yechish: Sistemaning cheksiz ko’p yechimga ega bo’lishlik sharti 2-dan 3 9 = 6 18 = k k + 1 ni olamiz. Bu yerdan k = 1 2 . Javob: 1 2 (C). 7. (01-2-15) k ning qanday qiymatida ½ 3x + (k − 1)y = k + 1 (k + 1)x + y = 3 tenglamalar sistemasi cheksiz ko’p yechimga ega bo’ladi? A) −1 B) −2 C) 0 D) 2 8. (98-3-24) k ning qanday qiymatlarida ½ (k 2 − k − 1)x + 2, 5y = 5 2x + y = −k sistemaning birorta ham yechimi bo’lmaydi? A) −2 B) −2 va 3 C) 3 D) 4 va 3 Yechish: Sistemaning yechimga ega bo’lmaslik sharti 3 − dan ½ k 2 − k − 1 = 5 k 6= −2 66 ni olamiz. Sistema 1− tenglamasining yechimlari k 1 = −2 va k 2 = 3 lardir. Ikkinchi munosabat k 6= −2 dan yechim faqat k = 3 bo’ladi. Javob: k = 3 (C). 9. (98-5-20) a ning qanday qiymatlarida ½ ax − y = 0 x + y = 10 tenglamalar sistemasi yechimga ega bo’lmaydi? A) −1 B) 2 C) 1 D) −2 10. (98-10-71) k ning qanday qiymatlarida ½ (k 2 + k + 1)x + 3y − 6 = 0 x + y + k = 0 sistema birorta ham yechimga ega bo’lmaydi? A) −2 B) 1 C) −2 va 1 D) 2 va 3 11. (02-5-10) m ning qanday qiymatlarida ½ mx + 2y + 4 = 0 2x + my − 8 = 0 tenglamalar sistemasi yechimga ega emas? A) 4 B) −4 C) 2 D) −2; 2 12. (99-2-17) a ning qanday qiymatida ½ 2x + ay = 2 ax + 2y = 3 tenglamalar sistemasi yechimga ega bo’lmaydi? A) 3 B) ±3 C) 4 D) ±2 13. (99-1-17) Tenglamalar sistemasi a ning qanday qiymatlarida yechimga ega emas? ½ a 2 x + 3y = 3 3x + y = 4 A) ±3 B) ±1 C) ± √ 3 D) 0 14. (03-10-30) a ning nechta qiymatida ½ (a − 2) Download 1.09 Mb. Do'stlaringiz bilan baham: |
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