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3 − 4 19. (03-7-50) Hisoblang. q 11 − 4 √ 7 A) √ 7 + 2 B) √ 7 − 2 C) √ 7 D) 2 − √ 7 20. (97-8-27) Hisoblang. q 11 − 6 √ 2 + q 11 + 6 √ 2 A) 8 B) 4 C) 3 D) 6 21. (97-12-27) Hisoblang. q 9 − 4 √ 2 − q 9 + 4 √ 2 A) 2 B) 3 C) −3 D) −2 22. (98-7-14) Hisoblang. ³q 3 − √ 5 + q 3 + √ 5 ´ 2 · 0, 5 −2 A) 38 B) 30 C) 40 D) 44 23. (99-10-14) Hisoblang. p 3 + 2 √ 2 + p 3 − 2 √ 2 4 √ 2 A) √ 2 4 B) 0, 5 C) √ 2 2 D) 0, 75 24. (03-1-1) Soddalashtiring. r ( π 2 − √ 3) 2 + r ( π 3 − √ 2) 2 − q 5 + 2 √ 6 A) √ 3 + √ 2 B) 5π 6 − 2( √ 2 + √ 3) C) 5π 6 D) −5π 6 43 3.1.2 Ildizli ifodalarni soddalashtirish Biz bu bandda arifmetik ildizning xossalari hamda qisqa ko’paytirish formulalaridan foydalanib misollar yechamiz. 1. (01-1-62) Ifodani a ≥ 0, 5 da soddalashtiring. √ a 2 − p a 2 + a + 0, 25 + p a 2 − a + 0, 25 A) a − 0, 25 B) a − 0, 5 C) a − 0, 75 D) a − 1 Yechish: a ≥ 0, 5 shartdan hamda a 2 + a + 0, 25 = (a + 0, 5) 2 , a 2 − a + 0, 25 = (a − 0, 5) 2 tengliklardan, 3.1-mavzudagi 6-qoidaga ko’ra √ a 2 − p a 2 + a + 0, 25 + p a 2 − a + 0, 25 = = a − (a + 0, 5) + (a − 0, 5) = a − 1 tenglikni olamiz. Javob: a − 1 (D). 2. (97-6-56) Quyidagi ifodani m = 15 va n = 3 √ 2 bo’lgandagi qiymatini hisoblang. ( √ m + n) p m − 2 √ m · n + n 2 m − n 2 A) 1 B) −1 C) −3 D) 0 3. (01-7-14) Agar a = (2 + √ 3) −1 va b = (2 − √ 3) −1 bo’lsa, (a + 1) −1 + (b + 1) −1 ning qiymatini hisoblang. A) 2 B) 0, 5 C) 2 √ 3 D) 1 4. (01-10-1) Agar a = √ 2 va b = 3 √ 3 bo’lsa, p a 2 − 2ab + b 2 + p a 2 + 2ab + b 2 ning qiymatini hisoblang. A) √ 8 B) 3 √ 12 C) √ 18 D) 3 √ 24 5. (02-6-8) x = 5 √ 6 va y = 6 √ 5 bo’lsa, p x 2 + 2xy + y 2 − p x 2 − 2xy + y 2 ning qiymatini hisoblang. A) √ 720 B) √ 700 C) √ 640 D) √ 600 6. (97-1-57) Ushbu (x + √ y) · p y − 2 · √ y · x + x 2 y − x 2 ifodani x = 2 √ 6 va y = 23 bo’lganda hisoblang. A) 1 B) −1 C) 1 2 D) − 1 2 7. (03-10-12) Agar x = 4 5 m bo’lsa, √ m + x + √ m − x √ m + x − √ m − x ning qiymatini toping. A) 2 B) 2m C) 4 D) −2 8. (03-10-15) Agar x < 0 bo’lsa, p x 2 − 12x + 36 − √ x 2 ni soddalashtiring. A) 6 B) −6 C) 6 − 2x D) 2x − 6 9. (01-11-7) Soddalashtiring. 3 a − √ a 2 − 3 + 3 a + √ a 2 − 3 A) 1, 5a B) 3a C) 2a D) 2, 5a Yechish: Birinchi kasr surat va maxraji a + √ a 2 − 3 ga, ikkinchisini esa a− √ a 2 − 3 ga ko’pay- tiramiz, natijada kasrlarning maxraji irratsional- likdan qutiladi va biz a + p a 2 − 3 + a − p a 2 − 3 = 2a ni olamiz. Javob: 2a (C). 10. (02-12-25) Soddalashtiring. 3 a − √ a 2 − 3 − 3 a + √ a 2 − 3 A) 3a B) 3 √ a 2 − 3 C) 6a D) 2 √ a 2 − 3 11. (02-11-12) Soddalashtiring. Ã 1 + √ x + x x √ x − 1 ! −1 − x 1 2 A) √ x + 1 B) 1 C) √ x − 1 D) −1 12. (02-12-13) Soddalashtiring. √ x + 1 x √ x + x + √ x : 1 √ x − x 2 + x A) 2x B) 2 C) 1 D) 2x − 1 13. (03-8-41) Kasrni qisqartiring. c − 2 √ c + 1 √ c − 1 A) √ c − 1 B) c − 1 C) c + 1 D) √ c + 1 14. (03-11-5) Soddalashtiring. a · Ã √ a + √ b 2b √ a ! −1 + b · Ã √ a + √ b 2a √ b ! −1 A) 2ab B) ab C) 4ab D) 0, 5ab 15. (99-1-15) Soddalashtiring. ³ 1 √ a + √ b − √ a + √ b a − b ´ · a − 2 √ a √ b + b 2 √ b A) √ b − √ a √ a + √ b B) √ a − √ b √ a + √ b C) √ b + √ a √ a − √ b D) √ a − √ b a + b 16. (00-1-20) Soddalashtiring. ³ 1 √ a + 1 + √ a + 1 √ a − √ a − 1 ´ ( √ a + 1− √ a − 1) A) 1 B) 2 C) 2 √ a D) 2 √ a − 1 44 3.2 n-darajali ildiz n natural son uchun n − darajasi a ga teng bo’lgan b son a sonning n − darajali ildizi deyiladi. n − dara- jali ildizdan b sonni topish a sondan n − darajali ildiz chiqarish deyiladi. a sondan n − darajali ildiz chiqarish quyidagicha belgilanadi: n √ a = b. (3.2) Bu yerda a ildiz ostidagi son, n − ildiz ko’rsatkichi deyiladi. (3.2) tenglik a va b sonlari orasida quyidagi bog’lanish borligini bildiradi: b n = a. Agar a ≥ 0 ix- tiyoriy son va n > 1 natural son bo’lsa, b n = a teng- likni qanoatlantiruvchi manfiy bo’lmagan b son mavjud bo’lib, u yagonadir. Bu b son a ≥ 0 sonning n − dara- jali arifmetik ildizi deyiladi. a sonning n − darajali ildizi n √ a yoki a 1 n ko’rinishida yoziladi. Kasr ko’rinishda- gi sonlar ratsional sonlar bo’lganligi uchun a m n ni rat- sional ko’psatkichli daraja ham deyiladi. Ko’pincha ildiz atamasi o’rniga radikal termini ham ishlatiladi. Ildiz ko’rsatkichi n = 2 bo’lgan ildiz, kvadrat ildiz deyiladi. 1 n = k kn tenglikdan n √ a = a 1 n = a k kn = kn √ a k tengliklar kelib chiqadi. Bu xossadan foydalanib har xil ildiz ko’rsatkichga ega bo’lgan ildizlarni bir xil ildiz ko’rsatkich orqali ifodalash mumkin. Masalan, 3 √ 2 va 4 √ 4 larni 12 √ 2 4 = 12 √ 16 va 12 √ 4 3 = 12 √ 64 shaklda yozish mumkin. 1. Agar ildiz ko’rsatkichi n − juft son va a > 0 bo’lsa, a son uchun ikkita n − darajali ildiz mavjud bo’ladi. Bu sonlar absolyut qiy- matlari teng bo’lgan o’zaro qarama-qarshi sonlardir. Masalan, √ 16 = ±4, 4 √ 81 = ±3. 2. Agar ildiz ko’rsatkichi n − toq son va a > 0 bo’lsa, a ning ildizi yagona va musbat bo’ladi. Masalan, 5 √ 243 = 3, 3 √ 1, 728 = 1, 2. 3. Agar ildiz ko’rsatkichi n − toq son va a < 0 bo’lsa, a ning ildizi yagona va manfiy bo’ladi. Masalan, 3 √ −8 = −2, 5 √ −243 = −3. 4. Agar ildiz ko’rsatkichi n − juft son va a < 0 bo’lsa, a sonning haqiqiy ildizi mavjud emas, chunki manfiy sonning juft darajasi musbat son bo’ladi. 5. Agar ildiz ko’rsatkichi n − natural son va a = 0 bo’lsa, a sonning ildizi nolga teng bo’ladi, ya’ni n √ 0 = 0, chunki 0 n = 0. n − darajali arifmetik ildiz quyidagi xossalarga ega. Bu yerda a > 0, b > 0 va n, m ∈ N sonlar. 1. a n m = m √ a n . 2. m √ a · b = m √ a · m √ b. 3. m r a b = m √ a m √ b . 4. a m √ b = m √ a m · b. 5. n p m √ a = nm √ a. 6. ( n √ a) m = n √ a m . 7. n √ a = nm √ a m 8. ( n √ a) n = a, a ≥ 0. 9. 2n+1 √ −a = − 2n+1 √ a. 10. ( a b ) −n = ( b a ) n . 1. (00-3-17) Soddalashtiring. a − a √ a 3 √ a 2 + 6 √ a 5 + a + 3 √ a 2 − a 3 √ a + √ a + 2 √ a A) 2 3 √ a B) 2 √ a C) 3 √ a + 2 √ a D) 0 Yechish: 6 √ a = x deb belgilaymiz. U holda a = x 6 , 3 √ a 2 = x 4 , √ a = x 3 , 3 √ a = x 2 bo’ladi. U holda berilgan ifoda x 6 − x 6 · x 3 x 4 + x 5 + x 6 + x 4 − x 6 x 2 + x 3 + 2x 3 ko’rinishga keladi. Uni soddalashtiramiz: x 6 (1 − x 3 ) x 4 (1 + x + x 2 ) + x 4 (1 − x 2 ) x 2 (1 + x) + 2x 3 = = x 2 (1 − x)(1 + x + x 2 ) 1 + x + x 2 + x 2 (1 − x)(1 + x) 1 + x +2x 3 = = x 2 (1 − x) + x 2 (1 − x) + 2x 3 = x 2 − x 3 + +x 2 − x 3 + 2x 3 = 2x 2 = 2 3 √ a. Javob: 2 3 √ a (A). 2. (99-5-5) Soddalashtiring. 27a + 1 9a 2 3 − 3 3 √ a + 1 − 27a − 1 9 3 √ a 2 + 3a 1 3 + 1 A) 3 √ a − 1 B) 1 C) 2 D) a + 1 3. (00-8-53) Soddalashtiring. a 3 4 − 36a 1 4 a 1 2 − 6a 1 4 A) 4 √ a−6 B) 4 √ a+6 C) √ a−6 D) √ a+6 4. (00-9-14) Soddalashtiring. 729a + 1 81 3 √ a 2 − 9a 1 3 + 1 − 729a − 1 81a 2 3 + 9 3 √ a + 1 A) 1 B) 2 C) 3 D) 9 5. (97-9-81) Soddalashtiring. 3 √ x 2 + 2 3 √ x + 1 x + 3 3 √ x 2 + 3 3 √ x + 1 − 1 3 √ x + 1 A) 1 B) 1 3 √ x + 1 C) 3 √ x D) 0 45 6. (98-5-17) Ifodani soddalashtiring, (a 1 2 − b 1 2 )(a + a 1 2 · b 1 2 + b) so’ng a va b lar daraja ko’rsatkichlarining yig’in- disini hisoblang. A) 2 B) 1 C) 4 D) 3 Yechish: Agar a 1 2 = x va b 1 2 = y desak, u holda a = x 2 , b = y 2 bo’ladi. U holda berilgan ifoda qisqa ko’paytirish formulasining 6-ga ko’ra (x − y)(x 2 + xy + y 2 ) = x 3 − y 3 = a 3 2 − b 3 2 ko’rinishga keladi. Endi a va b larning darajalar- ini qo’shamiz 3 2 + 3 2 = 3. Javob: 3 (D). 7. (99-7-19) Ifodani soddalashtirib, (a 1 2 + b 1 2 )(a − a 1 2 · b 1 2 + b) a va b asosli darajalar ko’rsatkichlarining yig’in- disini toping. A) 1 B) 4 C) 2 D) 3 8. (98-5-18) Soddalashtiring. (5b 1 4 + 10)(b 3 4 − 2b 1 2 ) b − 4b 1 2 A) 1 1 4 B) 1 5 C) 1 D) 5 9. (02-10-7) Soddalashtiring. Ã 9 a + 8 − a 1 3 + 2 a 2 3 − 2a 1 3 + 4 ! · a 4 3 + 8a 1 3 1 − a 2 3 + 5 − a 2 3 1 + a 1 3 A) 5 B) 1 1 − a C) 2 1 − a 2 3 D) 4 10. (01-5-5) Soddalashtiring. a − b a + b + 2 √ ab : a − 1 2 − b − 1 2 a − 1 2 + b − 1 2 A) −1 B) a + b C) 1 √ a + √ b D) ab a + b 11. (98-9-18) Agar n = 81 bo’lsa, 3 p n √ n qiymati qanchaga teng bo’ladi? A) 3 B) 6 C) 9 D) 4 Yechish: 4 va 5-xossalardan foydalanib 3 p n √ n = 3 p√ n 2 · n = 6 √ n 3 = √ n ni olamiz. Endi n = 81 desak, √ 81 = 9. Javob: 9 (C). 12. (02-12-44) Agar a = 729 bo’lsa, a 4 3 − 8a 1 3 a 2 3 + 2a 1 3 + 4 : ( 3 √ a − 2) ning qiymatini toping. A) 9 B) 6 C) 12 D) 15 13. (03-4-9) Agar x = 256 bo’lsa, x − 1 x 3 4 + x 1 2 · x 1 2 + x 1 4 x 1 2 + 1 · x 1 4 + 1 ning qiymatini toping. A) 14 B) 15 C) 16 D) 13 14. (03-4-28) a = 64 bo’lganda, a 4 3 − 8a 1 3 b a 2 3 + 2a 1 3 b 1 3 + 4b 2 3 : Ã 1 − 2b 1 3 a 1 3 ! − 4a 2 3 ning qiymatini hisoblang. A) −46 B) −48 C) −44 D) −50 15. (03-8-14) Quyida berilgan ifodaning x = √ 3− 3 √ 2 bo’lgandagi qiymatini toping. x 2 − 2x √ 3 − 3 √ 4 + 3 x − √ 3 A) √ 3 B) 3 √ 2 C) 1 D) 0 Yechish: Dastlab ifodani soddalashtiramiz, keyin x o’rniga berilgan qiymatini qo’yamiz. Berilgan kasr surati x 2 − 2x √ 3 − 3 √ 4 + 3 ni x 2 − 2x √ 3 + ( √ 3) 2 − 3 √ 2 2 = (x − √ 3) 2 − ( 3 √ 2) 2 shaklda yozib olamiz. Bu ifodaga ikki son kvadratlarining ayir- masi uchun formulani qo’llab (x − √ 3 − 3 √ 2)(x − √ 3 + 3 √ 2) ni olamiz. Agar x o’rniga √ 3 − 3 √ 2 ni qo’ysak ikkinchi qavs ichi nolga aylanadi. Shun- ing uchun ko’paytma nol bo’ladi. Javob: 0 (D). 16. (99-9-2) Agar a = 27 bo’lsa, Ã a − b 3 √ a − 3 √ b + 3 √ ab ! : ³ 3 √ a + 3 √ b ´ + + ³ 3 √ a 2 − 3 √ b 2 ´ : ³ 3 √ a + 3 √ b ´ ning qiymatini hisoblang. A) 4 B) 4 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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