abituriyentlar uchun variantlashtirilgan testlar to’plami

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ahmadjonov azamat unco academy axborotnoma 1999 1 a.99 abituriyentlar uchun variantlashtirilgan testlar to’plami o’qituvchi: ahmadjonov azamat foydalanuvchi: _______________________ ahmadjonov azamat unco academy axborotnoma 1999 2 1-variant 1.       2 61 16 81 1 :36 36 0,4 1 0, 4 0, 4                 ni hisoblang. a) 0,4 b) 0,(4) c) 14,4 d) 36 2. 20 209 7 ayirmaning oxirgi raqamini toping. a) 0 b) 7 c) 1 d) 3 3. 7 69 671 6673 66675    ni 6 ga bo’lishdagi qoldiqni toping. a) 4 b) 3 c) 1 d) 2 4.            8 2 4 8 20,8 0,2 1 0,2 1 0,2 1 0,2 1 5 ni hisoblang. a) 160,2 b) 1 c) 2 d) 3 5. institutdagi talabalarning …

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ahmadjonov azamat unco academy axborotnoma 1999 3 a) a b a b   b) a b a b   c) a b a b   d) b a a b   16.      2 4 8 128 2568 3 1 3 1 3 1 ... 3 1 3 1x      tenglamani yeching. a) 1 8 b) 1 2 c) -1 d) 1 17. 2 3 3 3 4 a x y x y       tenglamalar sistemasi a ning qanday qiymatlarida yechimga ega emas? a) 3 b) 1 c) 3 d)  18. 2 35 0x px   tenglamaning ildizlaridan biri 7 ga teng. ikkinchi ildizining va p ning qiymatini toping. a) -5; -2 b) -5; 2 c) 5; 2 d) 5; -2 19. 3 0 16 y x y x   …

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d) 2 2 2. aylananing o markazi to’g’ri burchakli abc uchburchakning ac gipotenuzasida yotadi. uchburchakning katetlari aylanaga urinadi. agar oc kesmaning uzunligi 4 ga, c nuqtadan cb katetning aylana bilan urinish nuqtasigacha bo’lgan masofa 3 ga teng bo’lsa, cb ni toping. a) 3 7 b) 7 c) 8 d) 6 3. abcd parallelogrammning perimetri 10 ga teng. abd uchburchakning perimetri 8 ga teng. bd diagonal uzunligini toping. a) 4 b) 2 c) 3 d) 3,5 4. trapetsiyaning yon tomoni uchta teng qismga bo’lingan, bo’linish nuqtalaridan ikkinchi yon tomoniga asosga parallel kesmalar o’tkazilgan. trapetsiyaning asoslari 2 va 5 ga teng bo’lsa, bu kesmalarning uzunliklarini toping. a) 2,5; 3,5 b) 3,5; 4,5 c) 2,5; 4 d) 3; 4 5.        1 2 3 41;2 , 3;4 , 4;3 , 0;5m m m m nutaqalardan qaysi birlari 2 2 25x y  tenglama bilan berilgan aylanada yotadi? …

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x n n z         14.    1 2 3 1ctg x tg x    tenglamaning  ;2  oraliqdagi yechimini toping. a) 2 b) 3 c) 4 d) 5 15. teng yonli trapetsiyaning katta asosi 2,7 ga, yon tomoni 1 ga, ular orasidagi burchak 600 ga teng. uning kichik asosini toping. a) 2,35 b) 1,7 c) 1,35 d) 2,7 3 16. 1x y  tenglama bilan berilgan to’g’ri chiziqqa parallel to’g’ri chiziqni toping. a) 1y x  b) 2x y  c) 1y x  d) 2 2 3 0x y   17. 2 2 3y x  to’g’ri chiziqning ox o’qi bilan hosil qilgan burchagini toping. a) 300 b) 600 c) 450 d) 1350 18.        2; 4 , 1;2 , 1; 2 2; 4a b c vad   …

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aholisining soni hosil bo’ladi. bu qanday raqam? a) 2 b) 3 c) 4 d) 6 ahmadjonov azamat unco academy axborotnoma 1999 6 29. 1 5 9 y x x     funksiyaning aniqlanish sohasiga tegishli barcha butun sonlar yig’indisini toping. a) 35 b) 28 c) 32 d) 30 30. 2 2, , . 2 2 25,m n k n m n nk    22 25.mn k    2 2 m n k  ni hisoblang. a) 1 b) 2 c) 5 d) 10 3-variant 1. 2 3 1,6 2,2 :1,4 11        hisoblang. a) 14 b) 140 c) 1,4 d) 0,14 2. 321 ga qanday eng kichik musbat sonni qo’shganda, yig’indi 6 ga qoldiqsiz bo’linadi? a) 0 b) 3 c) 5 d) 7 3. 3 3p n  son 1; 2; 3; 6; 9 va 18 ga qoldiqsiz …

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ahmadjonov azamat unco academy axborotnoma 1999 1 a.99 abituriyentlar uchun variantlashtirilgan testlar to’plami o’qituvchi: ahmadjonov azamat foydalanuvchi: _______________________ ahmadjonov azamat unco academy axborotnoma 1999 2 1-variant 1.       2 61 16 81 1 :36 36 0,4 1 0, 4 0, 4                 ni hisoblang. a) 0,4 b) 0,(4) c) 14,4 d) 36 2. 20 209 7 ayirmaning oxirgi raqamini toping. a) 0 b) 7 c) 1 d) 3 3. 7 69 671 6673 66675    ni 6 ga bo’lishdagi qoldiqni toping. a) 4 b) 3 c) 1 d) 2 4.         …

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