Zbirka Zadataka Za Prijemni Iz Matematike VA

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    R

    N = {1, 2, 3, . . .} Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}

    Q = mn |m ∈Z , n ∈N I = R \ Q

    a, b ∈R a < b (a, b) =

    { x

    ∈ R

    | a < x < b

    }

    (a, b] = {x ∈R |a < x ≤ b} [a, b] = {x ∈R |a ≤ x ≤ b} [a, b) = {x ∈R |a ≤ x < b} (−∞, + ∞) = R

    a b

    a b

    a b

    a b

    a) б )

    в ) г)

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    c1c2 . . . c p

    a1a2 · · ·am , b1b2 . . . bn c1c2 . . . c pc1c2 . . . c p . . . =

    = a1a2 · · ·a m , b1b2 . . . bn (c1c2 . . . c p), a i , b j , ck ∈ {0, 1, 2, . . . , 9}.

    3 10 = 0 , 3;

    121 10 = 12, 1 ;

    7 100 = 0 , 07;

    137 10000 = 0 , 0137

    1 720 : 2, 7 + 2 , 7 : 1, 35 + 4, 2 −1 340 ·0, 4 : 212 334 : 7

    1 2 −5, 25 : 10 + 12 − 25 : 0, 2

    1 720 : 2, 7 + 2 , 7 : 1, 35 + 4, 2 −1 340 ·0, 4 : 212 = = 2720 · 1027 + 270100 · 100135 + 4210 − 4340 · 410 · 25 = = 12 + 2 +

    168−4340 · 425 = = 52 + 12540 · 425 = 52 + 12 = 3 .

    334 : 7 1 2 −5, 25 : 10 + 12 − 25 : 0, 2 =

    = 154 : 15

    2 − 214 : 212 + 110 : 210 = = 12 − 12 + 12 = 12

    x

    x 0, 0016 : 0, 012 + 0, 7

    = 45425 : 14

    7 5 + 0 , 8

    1, 2 : 0, 375

    − 0, 2

    .

    x 0,0016:0 ,012+0 ,7 =

    4 5425 :14 7 5 +0 ,8

    1,2:0,375−0,2 x

    16 10000 :

    12 1000 +

    7 10

    = 154 25 :

    77 5 +

    8 10

    12 10 :

    375 1000 − 210

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    x 16

    10000 ·100012 + 710 =

    154 25 · 577 + 81012

    10 ·1000375 −15 x

    2 15 +

    7 10

    = 25 + 456 5 ·83 −15

    x 2

    15 + 7 10

    = 2 5 +

    4 5

    6 5 ·83 −15

    x 5 6

    = 6 5 3

    x = 13 .

    1, 75 : 23 −1, 75 : 712 17 80 −0, 0325 : 400

    : (6, 79 : 0, 7 + 0 , 3) .

    40 730 −38 512 : 10, 9 + 78 − 730 ·1 911 ·4, 2 0, 08

    .

    1 7 20

    : 2, 7 + 2 , 7 : 1, 35 + 0, 4 : 2 1 2 · 4, 2 −1

    3 40

    .

    7 2

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    sgn x = 1, x > 0 0, x = 0

    −1, x < 0 |x| =

    x, x ≥ 0 −x, x < 0

    ,

    x

    y

    0 -1

    +1

    y = sgn x

    y = sgn x

    x

    y

    0

    y=| x |

    y = |x|

    |x| = x · sgn x, n√ xn = x, n = 2k + 1 ,

    |x|, n = 2k, k

    ∈ N,

    a2 −b2 = ( a + b)(a −b) (a + b)2 = a2 + 2 ab + b2 (a −b)2 = a2 −2ab + b2 (a + b)3 = a3 + 3 a 2b + 3ab2 + b3 (a −b)3 = a3 −3a2b + 3ab2 −b3 a3 −b3 = ( a −b)(a 2 + ab + b2) a3 + b3 = ( a + b)(a2 −ab + b2) (a + b + c)2 = a2 + b2 + c2 + 2 ab + 2ac + 2bc

    a, b > 0 k ,m,n ∈Z am ·an = am + n , ab

    n = a n

    bn , n

    a b =

    n√ a n√ b ,

    a m a n = a

    m

    − n , a

    0 = 1 ,

    n

    √ a m

    = nk

    √ a mk

    , (am )n = am ·n , a−n = 1a n ,

    n√ am = nk√ a mk , n√ am = a mn , n√ a ·b = n√ a ·

    n√ b, m n√ a = n

    m√ a = nm√ a.

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    13 3 −1 : 0, 3 −

    5 : 1765 125 · 17 + 211

    −34 .

    ( 13 3 −1:0,3)

    −5 :

    176 5

    1 25 ·( 17 + 211 ) − 3 4

    =

    = ( 13 3 −103 )−

    5

    · 5176 7 5 ·2577

    −34 =

    = 5 176

    5 11

    −34 = 11176 − 3 4 =

    = 16 3 4 = (24)

    1 4

    3 =

    = 2 3 = 8 .

    38 ·9−2 ·54 + 9 ·125 · 15 − 1

    (3 ·5) 4

    ·3−3 : 5.

    38 ·9−2·54 +9 ·125·( 15 )− 1

    (3·5) 4·3−3 : 5 =

    = 38

    · 3−4

    · 54 +3 2

    · 53

    · 5

    (3·5) 4·3−3 : 5 = = 3

    4·54 +3 2 ·543·55 = = 3

    2·54 (3 2 +1) 3·55 =

    = 3 2·54·103·55 =

    32 ·55 ·23·55 = 6 .

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    4−14 + 1 2−32

    −43 · 4−0,25 − 2√ 2

    −43 .

    4−14 + 1 2−

    3 2

    −43 · 4−0,25 − 2√ 2 −

    4 3 =

    = (22)− 1 4 + 2

    3 2

    −43 · (22)−

    1 4 − 2 ·2

    1 2

    −43 =

    22·(−14 ) + 2 3 2 ·(−43 )

    · 22·(−14 )

    − 21+

    1 2

    −43 =

    = 2−12 + 2−2 · 2− 1 2 − 2

    3 2

    −43 =

    = 2−12 + 2−2 · 2− 1 2 −2−2 =

    = 2−12 2

    −(2−2) 2 =

    = 2−1 −2−4 = 12 − 116 = 716 .

    15

    √ 6 + 1 +

    4

    √ 6 −2 − 12

    3 −√ 6 :

    1

    √ 6 + 11

    15√ 6+1 + 4√ 6−2 −

    12 3−√ 6

    : 1√ 6+11 =

    = 15(√ 6−1)

    6−1 + 4(√ 6+2)

    6−4 − 12(3+ √ 6)

    9−6 : √ 6−116−121 =

    = 3(√ 6 −1) + 2(√ 6 + 2) −4 3 + √ 6 · −115√ 6−11 = = √ 6 −11 · −115√ 6−11 = −115.

    √ 7 −√ 5√ 7 + √ 5 + √ 7 + √ 5√ 7 −√ 5

    .

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    √ 7−√ 5√ 7+ √ 5 + √ 7+ √ 5√ 7−√ 5

    = .

    = √ 7−√ 5√ 7+ √ 5 ·√ 7−√ 5√ 7−√ 5 + √ 7+ √ 5√ 7−√ 5 ·√ 7+ √ 5√ 7+ √ 5 = = (

    √ 7−√ 5) 2

    (√ 7)2−(√ 5) 2 + (

    √ 7+ √ 5)2 (√ 7)2−(√ 5)

    2 =

    = ( 7−2·√ 7√ 5+5 )+ (7+2 ·√ 7√ 5+5 )

    2 = = 242 = 12.

    √ 5 −√ 2 + √ 3√ 5 + √ 2 + √ 3 .

    √ 5−√ 2+ √ 3√ 5+ √ 2+ √ 3 = = √ 5+ √ 3−√ 2√ 5+ √ 3+ √ 2 ·

    √ 5+ √ 3−√ 2√ 5+ √ 3−√ 2 =

    = ( √ 5+ √ 3−√ 2)

    2

    (√ 5+ √ 3)2−(√ 2) 2

    = 5+3+2+2 √ 5·√ 3−2√ 5·√ 2−2√ 3·√ 25+2 √ 5·√ 3+3 −2 =

    = 10+2 √ 15−2√ 10−2√ 66+2 √ 15 = = 5+ √ 15−√ 10−√ 63+ √ 15 · 3−

    √ 15 3−√ 15

    = −2√ 15+2 √ 69−15 = −2√ 15+2 √ 6−6 = √ 15−√ 63 .

    1 √ 2 + 3√ 3 .

    1√ 2+ 3√ 3 = 1√ 2+ 3√ 3 ·

    √ 2− 3√ 3√ 2− 3√ 3 =

    = √ 2− 3√ 3

    2− 3√ 9 =

    = √ 2− 3

    √ 32− 3√ 9 · 4+2 3

    √ 9+ 3

    √ 814+2 3√ 9+ 3√ 81 = = (

    √ 2− 3√ 3)·(4+2 3√ 9+ 3√ 81) 8−9 =

    = ( 3√ 3 −√ 2) ·(4 + 2 3√ 9 + 3√ 81).

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    a2 + b2

    ab − a2

    ab −b2 +

    b2

    a2 −ab .

    a 2 + b2 ab − a

    2

    ab−b2 + b2 a 2−ab =

    = a 2 + b2 ab − a

    2

    b(a−b) + b2 a(a−b) =

    = ( a 2 + b2 )(a−b)−a 2 ·a+ b2·b

    ab(a−b) = = a

    3 + ab2−ba2−b3−a 3 + b3ab(a−b) =

    = ab2

    − ba2

    ab(a−b) = ab(b

    − a )

    ab(a−b) = −1, ab = 0 , a = b.

    (a −b)2 ab

    + 3 · a b −

    b a

    : a3 −b3

    a2b2 .

    (a−b)2ab + 3 · ab − ba : a 3−b3a 2 b2 =

    = a 2

    −2ab+ b 2

    +3 abab · a 2

    −b 2

    ab · a 2

    b 2

    a 3−b3 = = (

    a 2 + ab+ b2)·(a 2−b2) a 3−b3 =

    = ( a 2 + ab+ b2)·(a−b)( a+ b)

    (a 2 + ab+ b2 )( a−b) == a + b a = b, ab = 0 .

    2 3

    a 3√ b b

    8

    √ a 12

    + 1

    2

    √ a a

    8

    √ b 3

    : 4√ a + 4√ b ; a,b > 0.

    2 3 a

    3√ b b 8√ a 12 +

    1 2 √ aa 8√ b3 : 4√ a + 4√ b =

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    = ab

    1 3

    3 2

    ba 128 3 2

    + a 1 2

    ab 3 8

    2 : a

    1 4 + b

    1 4 =

    = a 3 2 b

    1 3 ·

    3 2

    b 3 2 a

    12 8 ·

    3 2

    + a a 2 b

    3 8 ·2

    : a 1 4 + b

    1 4 =

    = a 3 2 b

    1 2

    b 3 2 a

    9 4

    + a a 2 b

    3 4

    : a 1 4 + b

    1 4 =

    = 1 ba

    3 4

    + 1 ab

    3 4

    : a 1 4 + b

    1 4 =

    = a 1 4

    ba + b

    1 4

    ab : a 1 4 + b

    1 4 = 1ab .

    1 √ a + √ a + 1 +

    1 √ a −√ a −1

    : 1 + a + 1a −1 .