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}.

    310 = 0 , 3;

    12110 = 12, 1 ;

    7100 = 0 , 07;

    13710000 = 0 , 0137

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

    12 −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 : 712 −5, 25 : 10 + 12 − 25 : 0, 2 =

    = 154 : 15

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

    x

    x0, 0016 : 0, 012 + 0, 7

    = 45425 : 14

    75 + 0 , 8

    1, 2 : 0, 375

    −0, 2

    .

    x0,0016:0 ,012+0 ,7 =

    4 5425 :1475 +0 ,8

    1,2:0,375−0,2x

    1610000 :

    121000 +

    710

    =15425 :

    775 +

    810

    1210 :

    3751000 − 210

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    x16

    10000 ·100012 + 710=

    15425 · 577 + 81012

    10 ·1000375 −15x

    215 +

    710

    = 25 + 4565 ·83 −15

    x2

    15 + 710

    =25 +

    45

    65 ·83 −15

    x56

    =653

    x = 13 .

    1, 75 : 23 −1, 75 : 7121780 −0, 0325 : 400

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

    40 730 −38 512 : 10, 9 + 78 − 730 ·1 911 ·4, 20, 08

    .

    1 720

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

    340

    .

    72

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    sgn x =1, x > 00, 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 = an

    bn , n

    ab =

    n√ an√ b ,

    a ma n = a

    m

    −n, a

    0= 1 ,

    n

    √ am

    = nk

    √ amk

    ,(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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    133 −1 : 0, 3 −

    5 : 1765125 · 17 + 211

    −34.

    (133 −1:0,3)

    −5:

    1765

    1 25 ·( 17 + 211 ) −34

    =

    = (133 −103 )−

    5

    · 517675 ·2577

    −34=

    = 5176

    511

    −34 = 11176 −34 =

    = 1634 = (24)

    14

    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 =

    = 32·54·103·55 =

    32 ·55 ·23·55 = 6.

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    4−14 + 12−32

    −43· 4−0,25 − 2√ 2

    −43 .

    4−14 + 12−

    32

    −43· 4−0,25 − 2√ 2 −

    43 =

    = (22)−14 + 2

    32

    −43· (22)−

    14 − 2 ·2

    12

    −43 =

    22·(−14 ) + 232 ·(−43 )

    ·22·(−14 )

    −21+

    12

    −43 =

    = 2−12 + 2−2 · 2−12 − 2

    32

    −43 =

    = 2−12 + 2−2 · 2−12 −2−2 =

    = 2−122

    −(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 −

    123−√ 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−

    √ 153−√ 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 + b2ab − a

    2

    ab−b2 + b2a 2−ab =

    = a2 + b2ab − a

    2

    b(a−b) + b2a(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)2ab

    + 3 ·ab −

    ba

    : a3 −b3

    a2b2 .

    (a−b)2ab + 3 · ab − ba : a3−b3a 2 b2 =

    = a2

    −2ab+ b2

    +3 abab · a2

    −b2

    ab · a2

    b2

    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 .

    23

    a 3√ bb

    8

    √ a12

    + 1

    2

    √ aa

    8

    √ b3

    : 4√ a + 4√ b ; a,b > 0.

    23 a

    3√ bb 8√ a 12 +

    12 √ aa 8√ b3 : 4√ a + 4√ b =

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    =ab

    13

    32

    ba 12832

    + a12

    ab38

    2: a

    14 + b

    14 =

    = a32 b

    13 ·

    32

    b32 a

    128 ·

    32

    + aa 2 b

    38 ·2

    : a14 + b

    14 =

    = a32 b

    12

    b32 a

    94

    + aa 2 b

    34

    : a14 + b

    14 =

    = 1ba

    34

    + 1ab

    34

    : a14 + b

    14 =

    = a14

    ba + b

    14

    ab : a14 + b

    14 = 1ab .

    1√ a + √ a + 1 +

    1√ a −√ a −1

    : 1 + a + 1a −1 . a ≥ 0, a +1 ≥ 0, a−1 ≥ 0 a = 1 a > 1.

    1√ a+ √ a +1 + 1√ a−√ a−1

    : 1 + a+1a−1 == √ a−√ a+1a−(a+1) + √ a+ √ a−1a−(a−1) : √ a−1+ √ a +1√ a−1 = √ a + 1 −√ a + √ a + √ a −1 :

    √ a−1+ √ a +1√ a−1=

    = √ a + 1 + √ a −1 · √ a

    −1

    √ a−1+ √ a+1 == √ a −1

    a3 −b3a + b− aba+ b −

    a3 + b3

    a −b + aba−b.

    a 3−b3( a + b) 2 −aba + b − a3 + b3( a −b) 2 + aba −b

    =

    = (a+ b)(a 3

    −b3 )

    a 2 + ab+ b2 − (a

    −b)(a 3 + b3)

    a 2−ab+ b2 ==

    (a+ b)( a−b)(a 2 + ab+ b2 )a 2 + ab+ b2 −

    (a−b)( a+ b)(a 2−ab+ b2 )a 2−ab+ b2 =

    = a2 −b2 −(a 2 −b2) = 0 , a = b, a = −b

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    12

    −8 ·16−2 + 2 −3 (0, 2)6

    ·50 + 13 −2

    ·(81−2)14 .

    3.

    1, 7 · (4,5·1 23 +3 ,75) 7135

    59 − 0, 5 + 13 − 512 .

    11784 .

    (16−2)−2 : 16(−2)−2 : 16−2−2 . 164

    √ 3 − 952

    − 3 85 3 −√ 3

    2.

    125

    1a 2 +

    1b2 · a

    3−b3a 2 + b2 :a 2 + b2

    ab + 1

    a−bab , ab = 0

    13√ 2+ 3√ 3 .

    = 153√ 4 − 3√ 6 + 3√ 9 .

    23

    10 2−5 5√ 340

    3−2 4√ 220 .

    a−8√ ab+4 ba−2

    4√ ab−2√ b + 3√ b 4√ a + 4√ b .

    1, a ≥ 0, b ≥ 0.

    1a 2 +

    1b2

    a 3−b3a 2 + b2 :a 2 + b2

    ab + 1

    a−bab , ab = 0 .

    1a+ 1b+ c

    : 1a+ 1b − 1b(abc+ a+ c) . 1, b = 0 , b+ c = 0 , ab = −1, a (b+ c) = −1, b(abc+ a+ c) = 0 .

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    D f ⊂ R D f ⊂R f R ×R

    (∀x ∈ D f ) ∃1y ∈ D f y = f (x), f : D f → D f ∃1

    D f f D f

    f

    f

    g

    D f = D g

    (∀x ∈ D f ) f (x) = g(x). f : D f → D f

    (∀x1, x 2 ∈ D f ) (x1 = x2 ⇒ f (x1) = f (x2)) . f : D f → D f

    y

    ∈ D f (

    x

    ∈ D f ) y = f (x).

    f : D f → D f

    A B C f : B → C, g : A → B f ◦g f g A C

    (∀x ∈ A) ( f ◦g)(x) = f (g(x)) . f : D f

    → D f f −1 : D f

    → D f

    (∀x ∈ D f ) f −1 ◦ f (x) = x ∀y ∈ D f f ◦f −1 (y) = y.

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    f 1(x) = x f 2(x) = x2x f 3(x) = √ x2 f 4(x) =(√ x)2

    f 1(x) = x x ∈R f 2(x) = x x ∈R \ {0} f 3(x) = |x| x ∈R f 4(x) = x x ∈ [0, ∞)

    f 1(x) = 2log 2 x f 2(x) = log 2 x2 f 3(x) =

    2log2 |x|

    f 4(x) = 2logx 2

    f 1(x) = 2 log2 x x ∈ (0, ∞) f 2(x) = 2 log2 |x| x ∈R \ {0} f 3(x) = 2 log2 |x| x ∈R \ {0} f 4(x) = 2 log2 x x ∈ (0, 1)∪(1, ∞)

    f 1(x) = f 2(x) = f 3(x) = f 4(x) = f 1(x) f 1(x) = eln x f 2(x) = x

    2

    x f 3(x) = √ x2 f 4(x) = ln( ex )

    f 1(x) = x x ∈ (0, ∞) f 2(x) = x x ∈R \ {0} f 3(x) = |x| x ∈R f 4(x) = x x ∈R

    f xx+1 = ( x −1)2 f (3)

    x xx+1 = 3 .

    x = −32

    f (3) = ( −32 −1)

    2

    = 6 .25

    xx+1 = t.

    x = t1−t

    f (t) = ( t1−t −1)2 =2t−11−t

    2 f (3) = ( 2·3−11−3 )

    2 = 6 .25

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    f x+12x−1 = x2008 −2x2007 + 1 f (f (2))

    x x+12x−1 = 2

    x = 1 f (2) = 1 2008−2·12007 +1 = 0 x x+12x−1 = 0

    x = −1 f (0) = ( −1)2008 −2 ·(−1)2007 + 1 = 4 f (f (2)) = f (0) = 4 f (2x −1) = x f (f (x))

    2x − 1 = t x = t+12 f (t ) = t+12

    f (f (x)) = f ( x+12 ) =x +1

    2 +12 = x+34

    f (x) = 5x+32x−5 g(x) = 1−x1+ x h(x) =

    3−x2+ x

    f (x) y = 5x+32x−5

    x 2x −5 2xy − 5y = 5x + 3 x = 3+5 y2y−5

    f −1(x) = f (x) g(x)

    y = 1−x1+ x

    x 1 + x y + xy = 1 −x x = 1−y1+ y

    g−1(x) = g(x) h(x)

    y = 3−x2+ x x 2 + x 2y + xy = 3 −x x = 3−2y1+ y

    h−1(x) = h(x) f (x) + 2 f (1 −x) = x x ∈R f (x)

    1

    −x = t x = 1

    −t

    f (1 −t) + 2 f (t) = 1 −t

    f (1 − t) = 1 − t − 2f (t) f (x) + 2 (1 −x −2f (x)) = x f (x) = 2−3x3 f (x) = x(x−1)( x−2)( x−3)( x−4)x−2−|x−2|

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    x −2 − |x −2| = 0 x ≥ 2

    0 = 0

    x < 2

    2(x−2) = 0

    x = 2

    D f = (−∞, 2) {0, 1, 2, 3, 4} x ∈ {0, 1}

    f (x) = 2 x −x2 f (f (f (1 −x)))

    f (f (f (1 −x))) == f (f (2(1 −x) −(1 −x)2)) == f (f (1 −x2)) == f (2(1

    −x2)

    −(1

    −x2)2) =

    = 2(1 −x4) −(1 −x4)2 == 1 −x8 f (x) = √ 1 −x2 g(x) = sin x

    6g f f −π4

    + f g −π4

    .

    6g f f

    −π4 + f g

    −π4 =

    = 6 g f 1 − −π42 + f sin −π4 =

    = 6 g f 1 − π216 + f −√ 22 == 6 g 1 − 1 − π216 + 1 − 24 == 6 sin π4 +

    √ 22 =

    7√ 22

    f g g (f (x)) = x2

    (x) = log16

    x f −32 + f (−1)

    g−1 (x) = 16 x f (x) = g−1 x2 = 16x2 = 4x

    f −32 + f (−1) = 4−32 + 4−1 = 18 +

    14 =

    38

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    f 1(x) = sin x f 2(x) = cos x · tg x f 3(x) = 1−cos2 x2 f 4(x) = |sin x|

    f 1(x) = f 2(x) = f 3(x) = f 4(x) = f 1(x) f 1(x) = sin 2 x + cos2 x f 2(x) = xx f 3(x) =√ x2

    x f 4(x) = log x x

    f x−1x+1 = x+2x+1 f (3)

    0

    f (3x + 2) = 2 x −1 f (f (x))

    x−89

    f (x) + 3 f 1x = x + 1 x ∈R \ {0} f (x)

    3+2 x−x28x

    f (x) = x(x−1)( x−2)( x−3)( x+4)log(2−x)

    {−4, 0}. f g g (f (x)) = 2 x (x) = 3 x + 1

    f (−1) + f (−2)

    −6 f (x) = 1−2x2+ x g(x) =

    x+13−x

    g (f (x)) + f (g (−x))

    2(7 x2 +8 x+13)5(x+1)( x+7)

    f (x) = 2x−14−x

    x ∈R \ {4} f : (−∞, 4)∪(4, ∞)

    1−1−→na (−∞, −2)∪(−2, ∞);f −1(x) = 4x+1x+2

    f (x) = x2 −2x f : (−∞, 1]

    1−1−→na [−1, ∞) f −1(x) = 1 −√ x + 1

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    P n (x) = an xn + a n−1xn−1 + . . . + a1x + a 0, an = 0 ,x ∈C n− an , a n−1, . . . , a 0 ∈C

    a0 = a1 = a2 = ... = an = 0 a 0 = 0 ∧ n = 0

    an = 1

    n ≥ 1

    P n (x) = an xn

    + an−1xn

    −1

    + .... + a2x2

    + a1x + a0, c P n (c) = 0

    x1, x 2,....,x p ∈ C , ( p ≤ n)

    P n (x) = an (x −x1)k1 (x −x2)k2 ...... (x −x p)kp x ∈ C k1, k2, ....k p

    k1 + k2 + .... + k p = n k1 x1, k2 x2 k p x p

    P n (x) Qm (x) = 0 , m < n S n−m (x) Rk(x), 0 ≤ k < m

    P n (x) = Qm (x)·S n−m (x)+ R k(x) P n (x) x −a P n ( )

    c C P n (x) k c̄

    P n (x) P n (x) = an xn + an−1x

    n−1 + . . . + a 1x + a 0, an = 0

    x1 + x2 + . . . + xn = −a n −1a nx1 ·x2 + . . . + xn−1 ·xn = an −2a n

    x1 ·x2 ·. . . ·xn = (−1)n · a0a n

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    n = 3 P 3(x) = a3x3 + a2x2 + a1x + a0, a3 = 0x1 + x2 + x3 = −a 2a 3

    x1 ·x2 + x1 ·x3 + x2 ·x3 = a1a 3x1 ·x2 ·x3 = −a 0a 3

    n = 4 P 4(x) = a4x4 + a3x3 + a2x2 + a1x + a 0, a4 = 0

    x1 + x2 + x3 + x4 = −a 3a 4x1 ·x2 + x1 ·x3 + x1 ·x4 + x2 ·x3 + x2 ·x4 + x3 ·x4 = a2a 4

    x1 ·x2 ·x3 + x1 ·x2 ·x4 + x1 ·x3 ·x4 + x2 ·x3 ·x4 = −a 1a 4x1

    ·x2

    ·x3

    ·x4 = a0a 4

    2x4 −3x3 + 4 x2 −5x + 6 x2 −3x + 1 x3 −3x2 −x −1 3x2 −2x + 1

    (2x4 − 3x3 + 4 x2 − 5x + 6) : (x2 −3x + 1) = 2 x2 + 3 x + 11 25x−5x2−3x+1

    −(2x4

    − 6x3 + 2 x2)3x3 + 2 x2 − 5x + 6−(3x3 − 9x2 + 3 x)11x2− 8x + 6

    −(11x2− 33x + 11)25x − 5 2x4 −3x3 + 4 x2 −5x + 6 x2 −3x + 1

    2x2 + 3 x + 11 25x −5.

    (x3 − 3x2 − x − 1) : (3x2 −2x + 1) = 13 x − 79 + −269 x−29

    3x2−2x+1

    −(x3

    − 2

    3 x2 + 13 x)

    (−73 x2 − 43 x − 1)− (−73 x2 + 149 x − 79 )− 269 x − 29

    x3−3x2−x−1 3x2−2x +1 13 x − 79 −269 x − 29

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    4x5 +9 x3 +19 x +92

    x + 1

    x + 1 P (−1) = 4(−1)5 + 9( −1)3 + 19(−1) + 92 = 60

    a P (x) = x4 + ax 2 + x −6 x + 2

    P (x) x + 2 x1 = −2 P (−2) = 0 16 + 4a −2 −6 = 0

    a = −2

    P n (x)

    x−1

    3

    x + 3 −1 x2 + 2 x −3

    P n (1) = 3 P n (−3) = −1 P n (x) = ( x2 + 2 x −3) ·K n−2(x) + ax + b P n (1) = 0 ·K n−2(1) + a + b = 3 P n (−3) = 0 ·K n−2(−3)−3a + b = −1 a + b = 3 3a + b = 1

    a = 1 b = 2 R(x) = x + 2

    P n (

    x)

    x 2

    x2 + 1 2x x3 + x

    P n (0) = 2 P n (x) = ( x2 + 1) ·R n−2(x) −2x P n (x) = ( x3 + x) ·K n−3(x) + ax 2 + bx + c ax 2 + bx + c P n (0) =

    c = 2 P n (i) = a i2 + b · i + c = 2 · i a = 2 , b = 2 R (x) = 2 ·x2 + 2 ·x + 2

    a b Q(x) = x2 + 2 x −3 P (x) = x4

    −ax 3 + bx2 + 9

    Q(x) = x2 + 2 x −3 (x + 3) ·(x −1)

    P (−3) = 0 ⇒ 81 + 27a + 9 b + 9 = 0 ⇒ 3a + b = −10

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    P (1) = 0 ⇒ 1 −a + b + 9 = 0 ⇒ −a + b = −10

    P (−3) = 0P (1) = 0 ⇒ 3a + b = −10−a + b = −10 ⇒

    a = 0b = −10

    a = 0 b = −10 P (x) = x4 −10x2 + 9

    P (x)

    P (x) = x4

    −ax3

    + bx2

    + 9 = ( x2

    + 2 x −3) ·(x2

    + Ax −3) == x4 + Ax 3 + 2 x3 −6x2 + 2 Ax 2 −6x −3Ax −9.

    x3 A + 2 = a x2 −6 + 2A = b x −6 −3A = 0

    A = −2, a = 0 , b = −10. P (x) = ( x2 + 2 x −3) ·(x2 −2x −3) = x4 −10x2 + 9

    x1, x

    2, x

    3

    125x3

    − 64 = 0

    x1 ·x2 ·x3 −(x1 + x2 + x3) x1 + x2 + x3 = −a 2a 3 = 0

    x1 ·x2 ·x3 = −a 0a 3 = 64125 x1 ·x2 ·x3 −(x1 + x2 + x3) = 64125 x3 + ax + b = 0 a b x1 = 1

    x2 = 2

    x1 + x2 + x3 = −a 2a 3 1 + 2 + x3 = 0 x3 = −3 x1 ·x2 ·x3 = −6

    x3 + 3 x2

    −4x

    −12

    x1+ x2+ x3 = −a 2a 3 = −3 x21 + x22 + x23 + 2 x1 ·x2 + 2 x1 ·x3 + 2 x2 ·x3 = 9

    x1 · x2 + x1 · x3 + x2 · x3 = a1a 3 = −4 x21 + x22 + x23 = 17

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    x3 −2x + b, b ∈ R

    1 + i, i2

    = −1

    1−i x1+ x2+ x3 = −a 2a 3 1 + i + 1 −i + x3 = 0 x3 = −2

    a b x4 + 2 x2 + ax + b x2 −1 x + 6

    x4 + 2 x2 + ax + b = ( x2

    −1)

    ·S 2(x) + x + 6

    x = 1 3 + a + b = 7 x = −1 −1 − a + b = 5 a = −1 b = 5

    x2008 + x2007 +1 x2 + 1

    x2008 + x2007 + 1 = ( x2 + 1) ·S 2006 (x) + ax + b x = i i2008 + i2007 +1 = ( i2 +1) ·S 2006 (i)+ a ·i + b

    2

    −i = a

    ·i + b a =

    −1, b = 2

    R (x) = −x + 2

    3x4 + 5 x3 −12x + 15 x −2 49

    a P (x) = x4+ ax 3−2x2−x+3 x −1 5

    a = 4

    P n (x) x + 1 5 x−2 2

    P n (x) x2 −x −2 R (x) = −x + 4

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    P n (x) x6 + 1

    x3

    + 2

    P n (x)

    x2 + 1 R (x) = −x + 2

    a, b Q(x) = x2 + x − 2 P (x) = x4 −5x2 + ax + b

    a = 0; b = 4

    x1, x 2, x 3 2x3 −x2 −4 = 0 x21 + x22 + x23

    14

    x3 − 4x2 + ax + b, a,b ∈ R 1 − 2i, i 2 = −1

    x3 = 2 , a = 9 , b = −10

    x1 = 2 , x2 = 2 , x3 = 3 −24 P 3(x) = 2 x3 −14x2 + 32 x −24

    x2015 +3 x2014 + x+3 x2 + 3 x

    R (x) = x + 3

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    y = n, (n ∈ R)

    x

    y

    0

    n

    y = nk = 0

    n > 0

    n < 0

    n = 0

    y = n

    y = kx + n, (k, n ∈ R), (k = 0) k

    x− k = tg α n y− x = −nk

    x

    y

    0

    k > 0 n > 0

    n = 0

    n < 0α

    y = kx + n, k > 0

    x

    y

    0

    k < 0

    n > 0

    n = 0

    n < 0

    α

    y = kx + n, k < 0

    ax = b x

    a = 0 x = ba

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    a = 0 b = 0 a = 0 b = 0

    y = mx −1 + 2 m m y A(0, 3) m

    3 = m

    ·0

    −1 + 2m 4 = 2m m = 2 .

    x

    y

    0

    3

    2

    3

    y = 2x + 3

    p T (6, −5) q : 2x + 3y + 5 = 0 .

    q y = −23 x − 53 . k = −23 .

    p y + 5 = −23 (x − 6) y = −23 x −1. M (−5, 4)

    P = 5

    M (−5, 4) y−4 = k(x +5) y = kx +5 k +4

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    x

    y

    0

    M (-5,4)

    -5

    4

    n

    m

    M (−5, 4) 5

    m n

    xm +

    yn = 1 , P =

    12 |mn |.

    M −5m +

    4n = 1

    5

    mn2 = 5

    m = 1 , n = 2 m = −52 , n = −4 2x + 5y −10 = 0 8x + 5y + 20 = 0 .

    3x + my −12 = 0 m

    3x12 +

    my12 = 1

    4 12m

    42 + 12m 2 = 5 16 + 144m 2 = 25 1449 = m2. m 2 = 16 m = ±4

    x−12

    + 3x−14

    = 2x−43

    + x+16

    NZS (2, 3, 4, 6) 6(x − 1) + 3(3x − 1) = 4(2x − 4) + 2(x + 1)

    15x − 9 = 10x − 14 5x = −5 x = −1

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    1x 2 +2 x+1 +

    2x+2 x2 + x3 =

    52x+2 x2

    1

    (x+1) 2 + 2x(x+1) 2 =5

    2(x+1) x = −1, x = 0 2x(x + 1) 2 2x + 4 = 5(1 + x) x = −13

    x − 1+ 34 x

    4 + 5−23 x

    4 = 3−x2

    3

    NZS (3, 4) = 12 12x − 3 1 + 34 x + 3 5 − 23 x = 4 3 − x2 .

    12x − 3 + 94 x + 15 − 103 x − 12 + 2x = 0 12x −94 x = 0 394 x = 0 . x = 0 x−2x+2 + x+2x−2 = 2

    (x−2)2 +( x+2) 2(x−2)( x+2) = 2

    2x2 +8x2−4 = 2

    2x2 +8x2−4 − 2

    x2−4x2−4 = 0.

    16x2−4 = 0

    |x + 2| − |2x −1| = 1

    |x + 2| = x + 2 , x ≥ −2−(x + 2) , x < −2 |2x − 1| = 2x −1, x ≥

    12

    −(2x −1), x < 12

    (−∞, −2) [−2, 12 ) [12 , + ∞) x

    −(x + 2) −(−(2x −1)) = 1 x = 4

    x x + 2 + 2 x − 1 = 1 x = 0

    x x + 2 − 2x + 1 = 1 x = 2

    {0, 2}.

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    |3x −1| − |2 −x| = 1

    |3x − 1| =3x −1, x ≥ 13−(3x −1), x < 13

    |2 − x| =2 −x, x ≤ 2−(2 −x), x > 2

    (−∞, 13 ) [13 , 2] (2, + ∞) x

    −(3x − 1) − (2 − x) = 1 x = −1

    x 3x − 1 − 2 + x = 1 x = 1

    x 3x − 1 + 2 − x = 1 x = 0

    {−1, 1}.

    2|x + 1| − |x −2| −3 = 0

    |x + 1| =x + 1 , x ≥ −1−(x + 1) , x < −1

    |x − 2| =x −2, x ≥ 2−(x −2), x < 2

    (−∞, −1) [−1, 2] (2, + ∞) x

    2(−(x +1)) −(−(x−2))−3 = 0 x = −7

    x 2(x + 1)

    −(

    −(x

    −2))

    −3 = 0

    x = 1 x

    2(x + 1) − (x − 2) − 3 = 0 x = −1 {−7, 1}.

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    x − 3y = 0 AB

    A(−1, 7)

    B (6, 8)

    π4

    C (3, −1). kx + ( k + 1) y − p = 0 k p

    M (2, 1)

    x + 2y −4 = 0

    5x−23 − 13x+17 = x−52 + x x = 1 .

    x+34 − 2x−12 = 7x−13 + 1 712 .

    x = 4 .

    1x3−27 +

    2x−3 =

    2x+1x2 +3 x+9 .

    x = −2. 1 − x1+ x1−x = x

    2. x ∈ ∅,

    |2x

    −4

    |+

    |x + 2

    | = 3 .

    x ∈ ∅, |2x −1| −x + 2 = |x −3|.

    x ∈ {0, 1}. |2x + 5| < 1

    x ∈ (−3, −2). |3x −2| > 4

    x

    ∈ (

    −∞,

    −23 )

    (2,

    ∞).

    2|x + 1 | < |x −2|+ 3 x + 1 x ∈ (−54 , ∞)

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    f (x) = ax 2 + bx + c a ,b,c ∈R a = 0 .

    f (x) = a x + b2a2

    − b2−4ac4a 2

    T (− b2a , −b2−4ac4a ).

    ax 2 + bx + c = 0 , a = 0 x1,2 = −b±√ b

    2−4ac2a D = b2 − 4ac

    D > 0

    D = 0

    D < 0

    x1 x2 ax 2 + bx + c = 0 ax 2 + bx + c = a(x −x1)(x −x2)

    x1 + x2 = −ba x1 ·x2 = ca

    m y = ( m + 2) x2 +(1 −m )x + m x = 2

    x = 2 x2 x

    m + 2 > 0 − b2a = − 1−m2(m +2) = 2 m = −3

    m + 2 > 0

    m f (x) = (1 +m )x2 −(4 + m)x + 8 −7 x = 3

    f (3) = (1 + m )32 −(4 + m) ·3 + 8 = −7 m = −2

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    a > 0 a < 0

    D > 0

    x

    y

    0 x 1 x 2 x

    y

    0 x 1 x 2

    D = 0

    x

    y

    0 x 1 = x 2 x

    y

    0 x 1 = x 2

    D < 0

    x

    y

    0

    x

    y

    0

    f (x) = ax 2 + bx + c

    x2

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    x

    y

    0-3 4

    f (x) = x2

    −x −12

    4x2 + mx + m 2 −15 = 0 m

    D = m2 −4 ·4 ·(m 2 −15) = −15m 2 + 240 = 0 m = ±4

    y = x2 + px + q p q

    −2 3

    x1 + x2 = −ba −2 + 3 = − p1 p = −1

    x1 ·x2 = ca −2 ·3 = q1 ⇔ q = −6 x2 −x −12 ≤ 0

    (x − 4)(x + 3) ≤ 0 (x + 3 ≥ 0∧x −4 ≤ 0)∨(x + 3 ≤ 0∧x −4 ≥ 0) −3 ≤ x ≤ 4 y = ( x−4)(x +3)

    −3 4 x2

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    y = ( x −4)(x + 3)

    x ∈ (−∞, −3) x = −3 x ∈ (−3, 4) x = 4 x ∈ (4, + ∞)x + 3 x −4

    (x −4)(x + 3) x ∈ [−3, 4]

    2x+3x−1 ≤ 0.

    x ∈ (−∞, −32 ) x = −

    32 x ∈ (−

    32 , 1) x = 1 x ∈ (1, + ∞)

    2x + 3 x −1

    (x −4)(x + 3) x ∈ [−32 , 1)

    x+1x−3 ≥ 2.

    x+1x−3 −2 ≥ 0

    x−7x−3 ≤ 0.

    x ∈ (−∞, 3) x = 3 x ∈ (3, 7) x = 7 x ∈ (7, + ∞)x −7 x −3

    (x −4)(x + 3) x ∈ (3, 7]

    (x+2)( x−1)( x−3)x(x+1) ≤ 0.

    f (x) (−∞, −2) −2 (−2, −1) −1 (−1, 0) 0 (0 , 1) 1 (1 , 3) (3, ∞)

    x + 2

    x −1 x −3

    x

    x + 1

    f (x)

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    x ∈ (−∞, −2]∪(−1, 0)∪[1, 3] m

    x2 −2mx + m2 −1 = 0 [−2, 4]

    x1,2 = 2m±√ 4m 2−4·1·(m 2−1)

    2 = m ±1 x1 = m −1 ≥ 2 x2 = m + 1 ≤ 4 −1 ≤ m ≤ 3

    y = ( m −1)x2 + ( m −4)x −(m + 1) m x = 1 − m−42(m−1) = 1 m = 2

    b y = 2x2 + bx −3 x = 54

    b = −5. (k−1)x2 −2kx −k +3 = 0 k

    k = 1∨k = −35 m

    (m −1)x2 + ( m −1)x −2 < 0 x ∈R m ∈ (−7, 1]

    x2 + 6 ≤ 5x x ∈ [2, 3]

    1x <

    12

    x ∈ (−∞, 0)∪(2, + ∞)

    3x−1

    < 22x+6 . x ∈ (−∞, −174 )∪(−52 , 1)

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    ax 4 + bx2 + c = 0 a(x2)2 + bx2 + c = 0

    x2 x2 = t at 2 + bt + c = 0

    ax 4 + bx3 + cx 2 + bx + a = 0 x2 x + 1x = t

    ax n ±b = 0 a, b n = 3 ax 3 ±b = 0 y = 3 ab y

    y3 ±1 = 0 (y ±1)(y2∓y + 1) = 0

    x4 −13x2 + 36 = 0 x2 = t t2 −13t + 36 = 0

    t1,2 = 13±√ 169−4·1·362 = 13±52

    t1 = 9 t2 = 4 x2 = 9 x2 = 4 x1 = 3 x2 = −3 x3 = 2 x4 = −2

    x4 −(m 2 + n2)x2 + m2n 2 = 0 x2 = t

    t2

    −(m 2 + n2)t + m2n 2 = 0

    t1,2 = +( m 2 + n 2 )±√ (m 2 + n 2 )2−4·1·m 2 n 2

    2 == +( m

    2 + n 2 )±√ m 4−2m 2 n 2 + n 42 == (m

    2 + n 2 )±√ (m 2−n 2 )22 =

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    = (m 2 + n 2 )±(m 2−n 2)

    2

    t1 = x2 = m 2 t2 = x2 = n 2 x1 = m x2 = −m x3 = n x4 = −n.

    6x4 −35x3 + 62 x2 −35x + 6 = 0 . x2 6x2 −35x + 62 − 35x + 6x2 = 0

    6(x2 + 1x2 ) −35(x + 1x ) + 6 = 0 x + 1x = t x2 +2+ 1x2 = t2 x2 + 1x2 = t2−2 6t2 −35t +50 = 0 . t = 103 ∨ t = 52 . x + 1x = 103 x + 1x = 52 . x2

    − 10

    3 x + 1 = 0

    x1 = 3 x2 = 13 . x2 − 52 x + 1 = 0

    x3 = 2 x4 = 12 .

    x4 −1 = 0

    (x − 1)(x + 1)( x2 + 1) = 0 , x1 = 1 x2 = −1 x3,4 = ±i

    x6 −729 = 0.

    (x3 −27)(x3 + 27) = 0 x3 −27 = 0 x3 + 27 = 0

    x1 = 3 x2,3 = 32 (−1±i√ 3) x4 = −3

    x5,6 = 32 (−1 ±i√ 3)

    x4 −34x2 + 225 = 0 x1,2 =

    ±5 x3,4 =

    ±3

    a2x4 −(a2 + b2)x2 + a2b2 = 0 x1,2 = ±ba x3,4 = ±a a = 0

    a = 0 b = 0 x = 0 a = b = 0 x ∈R .

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    4x4 + 2 x3 + 6 x2 + 3 x + 9 = 0

    x1,2 = 1

    ±i√ 52

    x3,4 = −3

    ±i√ 154

    x4 + 1 = 0 x1,2 =

    √ 2(−1±i )2 x3,4 =

    √ 2(1±i)2

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    x12 = 7

    (x2 + 1)12 = 4 (x + 1)

    12 = x + 2 √ x2 −5x + 2 = x −3

    f (x) = g(x) ⇔ (f (x) = ( g(x))2∧f (x) ≥ 0∧g(x) ≥ 0, f (x) < g (x) ⇔ (f (x) < (g(x))2∧f (x) ≥ 0∧g(x) ≥ 0,

    f (x) > g (x) ⇔ [(f (x) > (g(x))2∧g(x) ≥ 0]∨[g(x) < 0∧f (x) ≥ 0] .

    x −√ 11 + x2 = 11

    x −√ 11 + x2 = 11 ⇔⇔ −√ 11 + x2 = 11 −x ⇔⇔

    √ 11 + x2 = x −11 ⇔⇔ (11 + x

    2 = x2 −22x + 121)∧(11 + x2 ≥ 0)∧(x −11 ≥ 0) ⇔⇔ 22x = 110∧(11 + x

    2 ≥ 0)∧(x −11 ≥ 0) ⇔⇔

    x = 5

    (11 + x2

    ≥ 0)

    (x

    −11

    ≥ 0)

    5 − 11 ≥ 0

    5−√ 11 + 52 = 5−√ 36 =−1 = 11

    √ x −3 = 5 −x

    x −3 = (5 −x)2⇔ x2 −11x + 28 = 0 ⇔ x = 4∨x = 7 .

    √ 4 −3 = 5 − 4 √ 7 −3 = 5 −7

    √ x −9 −√ x −18 = 1

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    √ x −9 = 1 + √ x −18

    4 = √ x −18

    16 = x − 18 x = 34 √ 34 −9 −√ 34 −18 = 5 −4 = 1

    1 −√ x4 −x2 = x −1. x − 1 ≥ 0

    x ≥ 1 x4−x2 = x2(x2−1) ≥ 0 x2−1 ≥ 0 1 −√ x4 −x2 ≥ 0 1 −√ x4 −x2 = x2 − 2x + 1 √ x4 −x2 =x(2 −x) x2(x2 −1) = x2(2 −x)2

    x2(x2

    −1

    −4 + 4 x

    −x2) = 0 x2(4x

    −5) = 0

    x = 0 x = 54 .

    x = 54

    √ x + 5 > x −1

    x x+5 ≥ 0 x−1 < 0 x ∈ [−5, 1) x x −1 ≥ 0 x + 5 > (x −1)2

    x ≥ 1∧0 > x 2 −3x −4. x

    ∈ (

    −1, 4)

    x ≥ 1 x ∈ [1, 4)

    [−5, 1)∪[1, 4) = [−5, 4) √ x + 10 > x −2

    x x + 10 ≥ 0 x −2 < 0 x ∈ [−10, 2) x x −2 ≥ 0 x + 10 > (x −2)2

    x ≥ 2∧0 > x2

    −5x −6

    x ∈ (−1, 6) x ≥ 2 x ∈ [2, 6)

    [−10, 2)∪[2, 6) = [−10, 6)

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    √ x+22 = x −1. x = 2

    √ 2x + 1 = x −1 x = 4

    √ x −1 = x −3 x = 5

    √ x+1+ √ x−3√ 2x−2 = √ 3x−5√ x+1 −√ x−3

    . x = 3

    √ x + 4 < x −2. (2, 5).

    √ 2x2 −7x + 3 ≥ 0

    x ∈ (−∞, 12 ]∪[3, ∞)

    √ x2 −x −12 < x. x ∈ [3, ∞)

    (2x −4)12 −(x + 5)

    12 = 1

    x = 20.

    √ x + √ x + 9 = √ x + 1 + √ x + 4 x = 0 .

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    f (x) = ax , (a > 0, a = 1) D = R =( −∞, + ∞) a > 1

    0 < a < 1 a

    x

    y

    0

    y = a x

    a > 1

    1

    x

    y

    0

    y = a x

    0 < a < 1

    1

    y = ax a

    ax1 = ax2 x1 = x2

    a f (x) > b b ≤ 0 a f (x) f (x)

    a f (x) > b b > 0 a > 1 f (x) f (x) > loga b

    a f (x) > b b > 0 0 < a < 1 f (x)

    f (x) < loga b

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    y = 2x y = 12 x

    x

    y

    0

    y = 2 x

    1

    y = ( ) x12

    y = 2x 12

    x

    y = 3x + 1 y = |3x −1| y = 3x + 1

    y = 3x

    x

    y

    0

    y = 3 x+ 1

    1

    2

    y = 3x

    + 1

    y = |3x − 1| y = 3x x

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    x

    y

    0

    y = 3 x-1

    -1

    y = 3x

    −1

    x

    y

    0

    y = |3 x-1|

    1

    y =

    |3x

    −1

    |

    2x−1 = 4 5 9−3x = 127

    x+3

    3x −12 −2

    x +13 = 2

    x −23 + 3x −32

    0, 125 ·42x−3 =√ 28

    −x

    10 ·2x −4x = 16 2 ·3x+1 −5 ·9x−2 = 81

    2x−1 = 4 5 2x−1 = (2 2)5 2x−1 = 2 10

    x −1 = 10 x = 11 9−3x = 127

    x+3

    (32)−3x = (3 −3)x+3 3−6x = 3−3x−9 −6x = −3x −9 x = 3 .

    3x −12 − 2

    x +13 = 2

    x −23 + 3x −32 3

    x −12 − 3x −32 = 2

    x −23 + 2x +1

    3

    3x −32 (3−1) = 2

    x −23 (2+1) 2·3x −32 =

    3 ·2x −2

    3 3x −3

    2 ·3−1

    = 2x −3

    3 ·2−1

    3x −52 = 2

    x −53 312

    x−5 = 2 13x−5

    √ 33√ 2

    x−5 = 1 x −5 = 0 x = 5

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    0, 125 · 42x−3 =√ 28

    −x 2−3 · 24x−6 = 2−

    52

    −x

    24x−9 = 2 52 x 4x −9 = 52 x x = 6

    10·2x −22x = 16 2x = t 10t −t2 = 16 t1 = 8 t2 = 2

    2x = 2 3 2x = 2 1 x1 = 3 x2 = 1

    2 ·3x+1 −5 ·9x−2 = 81 2·3 ·3x −5 ·3−4 ·32x =81 6 · 81 · 3x − 5 · 32x − 812 = 0 3x = t

    486t −5t2

    −812

    = 0

    t1 = 81 t2 = 815 3x = 3 4 3x = 815 x1 = 4 x2 = 4 − log 5log 3

    5x −3x+1 > 2 (5x−1 −3x−2)

    14

    3x−2 ≤ 12 |x+1 |

    3x 3x > 0 x

    53

    x −3 > 2·15 · 53 x −29

    35

    53

    x > 3−29 35 53x > 259

    53

    x > 533

    53 > 1

    x > 3

    14

    3x−2 ≤ 12 |x+1 |

    12

    2 3x−2≤ 12 |

    x+1 |

    12

    6x−4 ≤ 12 |x+1 |

    12 < 1

    6x −4 ≥ |x + 1 | x + 1 x + 1

    ≥ 0

    6x − 4 ≥ x + 1 5x ≥ 5

    A = [1, ∞) x ≥ −1 x ∈ [1, ∞). x + 1 < 0 6x −4 ≥ x + 1 6x −4 ≤ −(x + 1) 5x ≥ 5∧x ≥ −1 7x ≥ 3∧x < −1 (x ≥ 1∧x ≥ −1) x ≥ 37 ∧x < −1 x ∈ [1, ∞).

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    y = 2x −2 y = 14 x + 1

    x

    y

    0

    y = 2 x - 2

    -1-2

    y = 2x −2

    x

    y

    0

    y = ( ) x+ 114

    1

    2

    y = 14x + 1

    x −7√ 32x+5 = 0 , 25·

    128x +17x −3

    x = 10

    2 ·3x+1 −4 ·3x−2 = 450 x = 4

    4√ x−2 + 16 = 10 ·2√ x−2

    x1 = 11 , x2 = 3

    23x ·3x −23x−1 ·3x+1 = −288 x = 2

    12

    |x+3 | ≤ 142x−3

    x ∈ (−∞, 3]

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    loga f (x) = b, (a > 0, a = 1) f (x) =a b loga f (x) = log a g(x), (a > 0, a = 1) f (x) = g(x)∧f (x) > 0∧g(x) > 0

    loga f (x) > b, (a > 0, a = 1) f (x) > a b a > 1 0 < f (x) < a b 0 < a < 1

    log2 3 ·log3 4 ·log4 5· · ·log7 8

    log 5 9√ 81 log30 3 = a log30 5 = b, log3 8

    log2 3 ·log3 4 ·log4 5· · ·log7 8 == log 2 3 · log2 4log2 3 ·

    log2 5log2 4 ·

    log2 6log2 5 ·

    log2 7log2 6 ·

    log2 8log2 7

    == log 2 8 = 3

    log 5 9√ 81 = 81 1log 5 9 = 9log9 5 2 = 5 2 = 25 .

    log30 8 = log30 23 = 3log 30 2 = 3 log303015

    = 3 (log30 30−log30 15)= 3 (1 −log30 3 ·5) = 3(1 −log30 3−log30 5) = 3(1 −a −b).

    log 12

    x = 3 x1+log 3 x = 9 log(x(x + 9)) = 1

    log3 x + log9 x + log81 x = 7 log4(x + 12) ·logx 2 = 1

    log49 x2 + log 7(x −1) = log7(log√ 3 3) 50log x ·160log x = 400

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    log12 x = 3

    x = 123 = 18

    x1+log 3 x = 9 (1 + log3 x) · log3 x = log3 32 = 2 log3 x = t

    (1 + t)t = 2 t2 + t −2 = 0 t1 = 1 t2 = −2 log3 x = 1 x = 3 log3 x = −2 x = 3−2 = 19

    log10(x(x + 9)) = log 10 10 x(x +9) = 10∧x(x +9) > 0

    x2 +9 x +10 = 0 x1 = 1 x2 = −10

    x(x + 9) > 0

    loga β b = 1β loga b log3 x+ 12 log3 x+

    14 log3 x = 7

    74 log3 x = 7

    log3 x = 4 x = 34 = 81∧x > 0.

    x > 0 log4(x+12) ·logx 2 = 1

    x+12 > 0 x > 0 x = 1 log4(x + 12)

    ·logx 2 = 1

    log4(x + 12) = log 2 x 12 log2(x + 12) = log 2 x log2(x + 12) = log 2 x2

    x + 12 = x2 x1 = 4 x2 = −3 x = 4

    log49 x2 + log 7(x −1) = log7(log√ 3 3) x = 0 x −1 > 0 log72 x2 + log 7(x −1) = log 7(log31/ 2 3)

    12 log7 x

    2 + log 7(x − 1) = log7(2log3 3) log7 x(x − 1) = log7 2 x > 0

    x2 − x = 2 x1 = −1 x2 = 2

    x = 2 50log x ·160log x = 400

    log 8000log x = log 400 log x·log 8000 = log 400 logx = log 400log 8000 = 2+2log23+3log2 = 23 x = 10

    23

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    log(2x + 3) ≤ 1

    log 133x

    −1

    x+2 > 1

    log(x + 1) < log(2x −1)

    log(2x +3) ≤ log 10 2x +3 ≤ 10∧2x +3 > 0

    x ≤ 72∧x > −32 x ∈ (−32 , 72 ]. 0 < 3x−1x+2 <

    13 0 <

    3x−1x+2 ∧3x−1x+2 < 13 x

    ∈ (

    −∞,

    −2)

    ( 13 , +

    ∞) = A 8x−5x+2 < 0

    x ∈−2, 58 = B A B x ∈13 , 58

    (x + 2 < 2x −1∧x + 2 > 0∧2x −1 > 0) 3 < x ∧x > −2∧x > 12 x ∈ (3, ∞) .

    log2(x −4) = 3

    x = 12

    log xlog x−log 3 = 2

    x = 9 , 1 + log2 x −1 = log x x = 10,

    2x2 = (2 x + 5) logx 4 ·log8 x x = 53

    logx (x + 2) > 2 x ∈ ∅

    log8(x2 −4x + 3) < 1 x ∈ (−1, 1)∪(3, 5)

    log√ 2 −x −log√ 4 −x2 + 3 log √ 2 + x < 2 x ∈ (−2, 2)

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    log12

    log8 x2−2xx−3

    < 0

    x ∈ (3, 4)∪(6, ∞)

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    α ABC sin α = ac =

    cos α = bc =

    tg α = ab =

    ctg α = ba =

    b

    a

    c

    α

    ABC

    ∠ pOq k1(O, r 1) k2(O, r 2)

    l1r 1 =

    l2r 2

    lr = 1 , l

    r 1

    π6

    π4

    π3

    π2 π

    3π2 2π

    30 45 60 90 180 270 360

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    p

    q

    O r 1

    r 2

    k1(O, r 1) k2(O, r 2).

    1

    x

    y

    0

    A(1,0)

    B(0,1)

    M

    y M

    x M

    C(1, tg x)

    D(ctg x, 1)

    q

    x

    sin x = xM cosx = yM tg x = xM yM , yM = 0 ctg x =yM xM

    , xM = 0 Oq

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    2π xM yM yM xM

    π

    y = sin x, x ∈R y = cos x, x ∈R 2π y = tg x, x = π2 + kπ, k ∈ Z y = ctg x,x = kπ, k ∈ Z π

    f (x) \ x 0 π6 π4 π3 π2 π 3π2 2πsin x 0 12

    √ 22

    √ 32 1 0 −1 0

    cos x 1√ 32

    √ 22

    12 0 −1 0 1tg x 0 √ 33 1 √ 3 − 0 − 0

    ctg x − √ 3 1√ 33 0 − 0 −

    x

    y

    0 x

    y

    0 x

    y

    0

    sin x cos x tg x, ctg x

    ++- -

    +

    +-- +

    +- -

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    sin(2kπ + α) = sin α, k ∈Z , sin(π + α ) = −sin α,cos(2kπ + α ) = cos α, k ∈

    Z, cos(π + α) = −cos α,sin( π2 −α ) = cos α, sin( 3π2 −α ) = −cos α,cos( π2 −α ) = sin α, cos(3π2 −α ) = −sin α,sin(π −α ) = sin α, sin(2π −α ) = −sin α,cos(π −α ) = −cos α, cos(2π −α ) = cos α,sin( π2 + α ) = cos α, sin( 3π2 −α ) = −cos α,cos( π2 + α) = −sin α, cos(3π2 −α ) = sin α.

    0−α sin(0−α ) =−sin α cos(0 − α ) = cos α

    sin2 α + cos2 α = 1

    tg α = sin αcos α , α = π2 + kπ, k ∈Z ctg α = cos αsin α , α = kπ, k ∈Z tg α ·ctg α = 1 , k = kπ2 , k ∈Z

    sin2

    α = sin2 α

    1 = sin2 α

    sin 2 α +cos 2 α = tg 2 αtg 2 α +1 , α =

    π2 + kπ, k ∈Z

    cos2 α = cos2 α

    1 = cos2 α

    sin 2 α +cos 2 α = 1

    tg 2 α +1 , α = π2 + kπ, k ∈Z

    sin(α + β ) = sin α cos β + cos α sin β

    sin(α −β ) = sin α cos β −cos α sin β cos(α + β ) = cos α cos β −sin α sin β cos(α −β ) = cos α cos β + sin α sin β tg( α + β ) = tg α +tg β1−tg α ·tg β

    α ,β ,α + β = π2 + kπ, k ∈Z tg( α −β ) = tg α−tg β1+tg α ·tg β α ,β ,α −β =

    π2 + kπ, k ∈Z

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    ctg( α + β ) = ctg α ·ctg β−1ctg α +ctg β ,α,β,α + β = kπ, k ∈Z ctg( α −β ) = ctg α ·ctg β +1ctg α−ctg β ,α,β,α + β = kπ, k ∈Z

    sin2α = 2 sin α cos α

    cos2α = cos2 α −sin2 α tg 2α = 2 tg α1−tg 2 α

    , α, 2α = π2 + kπ, k ∈Z ,ctg2α = ctg

    2 −12ctg α , α, 2α = kπ, k ∈Z . −

    2cos2 α2 = 1 + cos α cos α2 = ± 1+cos α2

    2sin2 α2 = 1 −cos α sin α2 = ± 1−cos α2 tg α2 = ± 1−cos α1+cos α , α = π + 2 kπ, k ∈Z ,ctg α2 = ± 1+cos α1−cos α , α = 2 kπ, k ∈Z .

    sin α + sin β = 2 sin α+ β2 cos α−β2

    sin α −sin β = 2 cos α+ β2 sin α −β2 cos α + cos β = 2 cos α + β2 cos

    α−β2

    cos α −cos β = 2 sin α + β2 sin α−β2 tg α ±tg β = sin( α±β )cos α cos β , α, β = π2 + kπ, k ∈Z ,ctg α

    ±ctg β = sin( β±α )sin α sin β , α, β

    = kπ, k

    ∈Z ,

    sin α cos β = 12 [sin(α + β ) + sin( α −β )] cos α cos β = 12 [cos(α + β ) + cos( α −β )] sin α sin β = −12 [cos(α + β ) −cos(α −β )]

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    x

    y

    0

    1

    -12

    -2

    3 22

    y = sin x

    y = sin x

    x

    y

    0 1

    -1

    2

    2

    y = arcsin x

    y = arcsin x

    f (x) \ x −1 −√ 32 −

    √ 22 −12 0 12

    √ 22

    √ 32 1

    arcsin x −π2 −π3 −π4 −π6 0 π6 π4 π3 π2arccos x π 5π6 3π4 2π3 π2 π3 π4 π6 0

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    x

    y

    0

    1

    -12

    -

    2

    3 22

    y = cos x

    y = cos x

    x

    y

    0 1-1

    2

    y = arccos x

    y = arccos x

    x

    y

    022

    3 2

    -

    y = tg x

    y = tg x

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    x

    y

    0

    y = arctg x2

    2

    y = arctg x

    x

    y

    022

    3 2

    -

    y = ctg x

    y = ctg x

    x

    y

    0

    y = arcctg x

    2

    y = arcctg x

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    f (x) \ x −√ 3 −1 −√ 33 0 √ 33 1 √ 3arctg x −π3 −π4 −π6 0 π6 π4 π3arcctg x 5π6 3π4 2π3 π2 π3 π4 π6

    sin x = a |a| ≤ 1 x = arcsin a + 2 kπ x = π−

    arcsin a + 2 kπ k

    ∈Z .

    cos x = a |a| ≤ 1 x = arccos a + 2kπ x = −arccos a + 2 kπ k ∈Z . tg x = a a

    x = arctg a + kπ k ∈Z . ctg x = a a

    x = arcctg a + kπ k ∈Z . sin ax ±sin bx = 0 cosax ±cos bx tg ax

    ±tg bx = 0 ctg ax

    ±ctg bx = 0

    a sin x + bcos x = 0 , a ,b = 0 x = π2 +kπ,k ∈ Z , cosx

    a sin x + bcos x = c, a, b, c = 0 , |c| < √ a2 + b2

    √ a2 + b2

    sin(ax + b) = sin( cx + d)

    ax + b = cx + d + 2 kπ x = d−ba−c +

    2kπa−c,

    k ∈ Z ax + b = π − (cx + d) + 2 mπ x = d+ ba + c +

    (2m +1) πa+ c, , m ∈Z .

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    a sin2 x+ bsin x+ c = 0 , a = 0 sin x = t

    (sin x > a, −1 ≤ a ≤ 1) arcsin a + 2 kπ < x < π −arcsin a + 2 kπ,.k ∈Z . (sin x < a, −1 ≤ a ≤ 1) π −arcsin a < x < 2π −arcsin a + 2 kπ,k ∈Z . (cos x > a, −1 ≤ a ≤ 1) −arccos a + 2 kπ < x < arccos a + 2 kπ,k

    ∈Z .

    (cos x > a, −1 ≤ a ≤ 1) arccos a +2 kπ < x < 2π −arccos a +2 kπ k ∈Z . (tg x > a, a ∈R ) arctg a + kπ < x < π2 + kπ, k ∈Z . (tg x < a, a ∈R ) π2 + kπ < x < arctg a + kπ, k ∈Z . (ctg x > a, a ∈R ) kπ < x < arcctg a + kπ, k ∈Z . (ctg x < a, a ∈R ) arcctg a + kπ < x < π + kπ, k ∈Z .

    sin 315 cos 1640 sin(−1320 )

    sin 315 = sin(360 −45 ) = −sin(45 ) = −−√ 22

    cos 1640 cos(4 ·360 + 200 ) = cos(180 + 20 ) = −cos 20 sin(

    −1320 ) =

    −sin 1320 =

    −sin(3

    ·360 + 240 ) =

    −sin 240 =

    = −sin(270 −30 ) = −(−cos30 ) = √ 32

    1+ctg 2 x·ctg xtg x+ctg x = 12 ctg x

    sin α +cos αsin α−cos α −

    1+2cos 2 αcos2 α (tg 2 α−1) =

    21+tg α ,

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    sin 4 α +2sin α cos α −cos4 αtg 2 α −1 = cos 2α

    cos α

    −sin α

    −cos3 α +sin 3 α

    2(sin 2 α +2cos 2 α−1) = sin α

    8cos 3 x−2sin 3 x+cos x2cos x−sin 3 x = −32 tg x = 2

    tg x ctg x

    1 + ctg 2 x ·ctg xtg x + ctg x

    = 1 + cos2xsin2 x · cos xsin x

    sin xcos x +

    cos xsin x

    =sin x sin2 x+cos 2 x cos x

    sin2 x sin xsin 2 x+cos 2 x

    sin x cos x

    =cos(2 x−x)sin2 x sin x

    1sin x cos x

    =

    = cos2 x sin x

    sin2x sin x =

    cos2 x2sin x cos x

    = 12 ·

    cos xsin x

    = 12

    ctg x,

    tg x

    sin α + cos αsin α −cos α −

    1 + 2 cos2 αcos2 α (tg 2 α −1)

    = sin α + cos αsin α −cos α −

    1 + 2 cos2 α

    cos2 α sin2 α

    cos2 α −1 =

    = sin α + cos αsin α −cos α −

    1 + 2 cos2 αsin2 α −cos2 α

    = (sin α + cos α )2

    sin2 α −cos2 α − 1 + 2 cos2 αsin2 α −cos2 α

    =

    = sin2 α + 2 sin α cos α + cos2 α −1 −2cos2 α

    sin2 α

    −cos2 α

    = 2sin α cos α −2cos2 α

    sin2 α

    −cos2 α

    =

    = 2cosα (sin α −cos α )(sin α −cos α )(sin α + cos α ) = 2cosα

    sin α + cos α = 2sin α

    cos α + 1 = 2

    tg α + 1,

    sin4 α + 2 sin α cos α −cos4 αtg 2α −1

    =sin2 α 2 −(cos2 α )

    2 + sin2 αtg 2α −1

    =

    =sin2 α + cos2 α · sin2 α −cos2 α + sin2 α

    tg 2α −1 =

    sin2 α −cos2 α + sin2 αtg 2α −1

    =

    = −cos2α + sin2αtg 2α −1

    = sin2α −cos2αsin2 α

    cos2 α −1 =

    sin2α −cos2αsin2 α−cos2 αcos2 α = cos 2α,

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    cos α −sin α −cos3α + sin3α2(sin2α + 2 cos2 α −1)

    = (cos α −cos3α ) + (sin3 α −sin α )

    2(sin2α + cos2α ) =

    = 2sin2α sin α + 2cos 2α sin α

    2(sin2α + cos2α ) =

    2sinα (sin2α + cos2α )2(sin2α + cos2α )

    = sin α,

    tg x = 2 sin x = 2cos x sin x

    8cos3 x −2 ·8cos3 x + cos x2cosx −8cos3 x

    = cosx −8cos3 x2(cos x −4cos3 x)

    = cosx(1 −8cos2 x)2cosx(1 −4cos2 x)

    =

    = 1−8cos2 x2(1 −4cos2 x) = sin2 x + cos2 x −8cos2 x2sin2 x + 2 cos2 x −8cos2 x

    = sin2 x −7cos2 x2sin2 x −6cos2 x.

    cos2 x tg 2 x−72 tg 2 x−6 =

    4−78−6 = −32 .

    x tg x = 34 , π < x ≤ 3π2

    sin x = ± tg x√ 1+tg 2 x sin x = ±34√ 1+ 916 = ±

    35

    π < x

    ≤ 3π

    2 sin x =

    −35

    cos x = −45 , ctg x = 43

    cos x + sin x = √ 2 sin x + cos x = sin x cos x + 1 sin2 x + sin 2 2x + sin 2 3x = 32

    1cos x = cos x + sin x

    cos2x + sin 2 x = cos x cos x cos π5 +sin x sin

    π5 =

    √ 32

    −π4 , 9π4 sin x

    2 + cos x = 1 .

    sin(x + π4 ) = 1 x + π4 = t sin t = 1

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    t = arcsin 1 + 2 kπ ∨t = π −arcsin1 + 2 kπ, k ∈Z arcsin 1 = π2 t = π2 +2 kπ∨

    t = π −π2 +2 kπ. π − π2 = π2 t = π2 + 2kπ, k ∈ Z .

    x + π4 = π2 + 2 kπ, k ∈Z x = π4 + 2 kπ, k ∈Z ;

    sin x−sin x cos x +cos x−1 = 0 sin x(1 − cosx) − (1 − cos x) = 0 . (sin x − 1)(1 − cosx) = 0 sin x −1 = 0 1 −cos x = 0 sin x = 1∨cos x = 1 x = π2 + 2kπ ∨x = 2kπ, k ∈Z ;

    1−cos2 x2 + 1−cos4 x2 +

    1−cos6 x2 = 3

    2 32 −

    12 (cos2x + cos4x + cos6x) =

    32

    cos2x + cos4x + cos6x = 0 2cos 6x+2 x2 cos

    6x−2x2 + cos 4x = 0 2cos4x cos2x + cos4x = 0

    cos4x(2cos2x + 1) = 0 cos4x = 0 cos2x = −12 4x = π2 + 2kπ 2x = ±2π3 + 2kπ k ∈Z x = π8 +

    kπ2 x =

    π3 + lπ x =

    π3 + mπ k , l ,m ∈Z

    cos x = 0 1cos x = cos x + sin x 1 − cos2 x − sin x cos x = 0 sin2 x − sin x cos x = 0

    sin x sin x = 0 sin x(sin x

    − cosx) = 0

    (sin x = 0 ∨ sin x = cos x) x = kπ∨x = π4 + lπ, k, l ∈Z cos2 x−sin2 x+sin 2 x = cos x cos2 x−cos x = 0

    cos x(cos x −1) = 0 (cos x = 0∨cos x = 1) x = π2 + kπ, k ∈Z∨x = 2lπ, l ∈Z ;

    cos(x − π5 ) =√ 32 . x − π5 = ±π6 + 2kπ,k ∈ Z x = π5 ± π6 + 2kπ, k ∈ Z (x = π30 + 2 kπ, k ∈Z∨x = 11π30 + 2 lπ, l ∈Z )

    π30 ,

    61π30

    11π30

    sin x2 + 1 −2sin2 x2 = 1 sin x2 − 2sin2 x2 = 0 . sin x2 1 −2sin x2 = 0 sin x2 = 0 sin x2 = 12 x2 = kπ, k ∈ Z

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    x = 2kπ, k ∈Z x2 =

    π6 + 2 lπ, l ∈Z∨

    x2 = π −

    π6 + 2mπ, m ∈Z

    x = π3 + 4 lπ, l ∈Z∨x = 5π3 + 4mπ, m ∈Z .

    2cosx + 1 ≥ 0 √ 3 tg x < 1

    sin x ≥ sin π7 sin x < cosx

    cosx

    ≥ −1

    2

    cos x = −12 x = ±2π3 + 2 kπ, k ∈ Z . cos x

    [−π, π ] −2π3 + 2kπ < x < 2π3 + 2kπ k ∈Z .

    x

    y

    0

    1

    -1

    -

    2

    y = cos x

    1

    2

    2 3

    2 3

    y = cos x

    tg x < 1√ 3

    tg x = 1

    √ 3

    x = π

    6 + kπ, k ∈ Z .

    tg x [−π2 , π2 ]

    −π2 + kπ ≤ x ≤ π6 + 2kπ k ∈Z

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    x

    y

    022

    y = tg x

    y = tg x

    sin x [0, 2π ]

    π7 + 2kπ ≤ x ≤ 6π7 + 2kπ k ∈Z

    x

    y

    0

    -1

    2

    - 22

    y = sin x

    7

    7

    y = sin x

    y = sin x y = cos x [−π, π ] sin x = cos x x = −3π4 x = π4

    −3π4 + 2kπ < x < π4 + 2kπ, k ∈Z

    sin −√ 22

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    x

    y

    0

    -1

    2

    -23

    2

    y = sin x

    y = cos x

    4

    3 4 2

    y = sin x y = cos x

    cos

    √ 22

    sin75 + sin15

    √ 62

    2·ctg α1+ctg 2 α = sin2α

    sin 2 α1+cos α · cos α1+cos 2 α = tg x2

    1−cos xsin x = sin x1+cos x = tg

    x2

    sin(α + β ) −sin(α −β ) sin α = 35 sin β = − 725 0 < α < π2 π < β < 32 π − 56125

    sin α2 cos α2 tg

    α2 cosα =

    119169

    sin α2 = ± 513 , cos α2 = ±1213 , tg α2 = 512

    sin3x = −√ 22

    x = −π4 + 2kπ ∨x = 5π4 + 2kπ, k ∈Z sin5x = sin 3x + sin x

    x = kπ∨x = ±π12 + kπ2 tg 4x = ctg 6 x

    x = π20 + kπ20 , k

    ∈Z

    √ 3sin x + cos x = √ 2 x = π12 + 2kπ ∨x =

    7π12 + 2 lπ, k, l ∈Z .

    sin x ≤ cos 3x

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    −π4 + 2 kπ ≤ x ≤ π8 + 2 kπ, k ∈ Z 5π8 + 2 kπ ≤ x ≤3π4 + 2kπ, k ∈Z

    9π8 + 2kπ ≤ x ≤

    13π8 + 2kπ, k ∈Z

    ctg x ≥ ctg π11 kπ < x ≤ π11 + kπ, k ∈Z

    2sin2 x −sin x −1 ≤ 0 −π6 + 2kπ ≤ x ≤ 7π6 + 2kπ, k ∈Z .

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    A B

    C

    ab

    c

    h c

    a, b, c

    α, β, γ

    α 1, β 1, γ 1

    ha , hb, hc

    s

    r

    R

    O = a + b + c = 2s

    P = a·h a2 = b·hb2 = c·h c2

    P = s(s −a)(s −b)(s −c) P = a·b·c4R = r ·s

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    asin α =

    bsin β =

    csin γ = 2R

    a2 = b2 + c2 −2bccos α,b2 = a2 + c2 −2ac cos β,c2 = a2 + b2 −2ab cos γ.

    A B

    C D

    d 2

    d 1

    .

    h a

    .

    h b

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    A B

    C D

    . .

    ..

    d

    d

    a

    a

    A B

    C D

    d 2

    d 1

    .

    h a

    .

    A Ba

    c d h

    .

    m

    m = a+ b2

    P = a+ b2 ·h = m ·h

    d1 d2 P = d1 d22

    a + c = b + d

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    a a

    bb

    d 1

    d 2.

    α + γ = 180 β + δ = 180

    n Dn = n(n−3)2

    n S n = ( n −2) ·180 n 360

    O = n ·a P = n · a·r2 a n r

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    A

    B

    C

    D

    a

    bc

    d

    r

    .

    R

    A B

    C

    D

    O

    O = 2rπ

    l = r

    ·π

    ·α

    180◦

    l = r · ϕ

    ϕ

    P = r 2π

    P = r2 ·π ·ϕ360 =

    12 ·l ·r = 12 ·r 2 ·ϕ

    P = 12 r2 πϕ

    180 −sinϕ = 12 r 2 (ϕ−sinϕ)

    r 1 r2 r1 > r 2 P = ( r 21 −r 22)π

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    O

    r 1

    r 1

    A

    B

    C

    D

    E

    F

    O

    b

    c

    a

    O = a + b + c = 132, a2 + b2 + c2 = 6050 c2 = a2 + b2

    2c2 = 6050 ⇒ c = 55

    a + b + 55 = 132a2 + b2 + 3025 = 6050

    a = 44 b = 33 c = 55

    5 12

    d(B, D ) = D (B, E ) = 5 d(A, D ) = d(A, F ) = 12 . d(O, d ) = d(O, E ) = d(O, F ) = r

    b = r + 12 ,a = r + 5 ,

    c = 17.

    (r + 5) 2 + ( r + 12) 2 = 17 2 r = 3 a = 8 b = 15

    a = 13 b = 14 c = 16 a b c

    P = s(s −a)(s −b)(s −c) s = a+ b+ c2 P = 21(21 −13)(21 −14)(21 −16) = 84 .

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    A B

    C

    . .

    .O

    ab

    c

    A B

    C D

    E a

    b

    d 1

    d 2

    P ABC = P AOC + P BOC P ABC = br2 +

    ar2

    r2 (13 + 14) = 84 r =

    569

    60

    d(A, B ) = a, d(A, C ) = b,d(B, D ) = d2, d(B, C ) = d1, ∠BAD = 60

    d21 + d22 = 2 (a 2 + b2) d21 : d :22= 19 : 7 d21 = 19

    7 d22 197 d

    22 + d22 = 267 d

    22 = 2( a2 + b2) a2 + b2 = 1317 d

    22

    d(A, E ) = bcos 60 = 12 b d(E, B ) = bsin60 = √ 32 b

    d(E, B ) = d22 − √ 32 b 2 = d22 − 34 b2 d(A, B ) = a = d(A, E )+ d(E, B ) = 12 b+ d22 − 34 b2 = 12 b+ 713 (a2 + b2) − 34 b2

    a − 12 b =

    7

    13 (a2 + b2) − 34 b2 6a2 − 13ab + 6b2 = 0 ,

    b2 6a

    2

    b2 − 13ab + 6 = 0 ab = t 6t2 −13t +6 = 0 t1 = 32 t2 = 23

    3 : 2

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    A B

    C D

    a

    ah

    r

    A O B

    C D

    R

    E

    P = π

    sin α = ha = 25

    a . P = a ·h = a ·a sin α = a2 2ra = 2ra = 8 r 2π

    r 2π = π r = 1 . 2ra = 8 a = 4 . sin α = 2ra =

    12 α = 30 .

    d(A, B ) = 21 d(C, D ) = 9 d(E, C ) = 8

    d(A, E ) = d(D, C ) + d(E, B ) == d(D, C ) + d(A,B )−d(D,C )2 =

    d(A,B )+ d(D,C )2 =

    21+92 = 15

    d(A, C ) = (d(A, E ))2 + ( d(E, C ))2 = √ 152 + 8 2 = 17 d(C, B ) = (d(E, B ))2 + ( d(E, C ))2 = 82 + 21−92 2 = 10 ABC

    R = abc4P = d(A,B )·d(A,C )·d(C,B )

    4·d( A,B ) ·d( C,E )2

    = d(A,C )·d(C,B )2d(C,E ) = 17·102·8 =

    858 .

    α = 120

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    A

    C

    B

    S

    R.

    O = 2Rπ3 + d(A, B ) d(C, B ) = R sin60 √ 32 R d(A, B ) = 2 d(C, B ) = √ 3R, O = 2πR3 + √ 3R = p. R = 3 p2π +3 √ 3

    P = R2 π 120

    360 − d(A,B )

    ·d(C,S )

    2 = R2 π

    3 −√ 3R

    ·R·cos 60

    2 = R2 π

    3 −√ 3R 2

    4

    P o = 3 p2π +3 √ 32

    π3 −

    √ 34 =

    3 p2 (4π−3√ 3)4(2π +3 √ 3)

    a = 12, b = 10

    r = 3 R = 12, 5.

    c = 4

    √ 3 + 1

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    90 60

    √ 2, 2

    R 2α

    P = R 2 ctg α

    60 O = 8 P = 2√ 3 2

    2

    a = 8 b = 2 c = 5

    30 . P = 12 2

    2

    a = 18 b = 8 c = 13

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    n n

    P = 2B + M V = B ·H B M H

    P = 2 ·(ab + bc + ac) V = a ·b ·c a b c D

    P = 6a2 V = a3

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    a

    b

    c D

    a

    a

    a D

    n n

    n n P = B + M

    V = 13 ·B ·H n

    a P = a 2√ 3 V = a

    3√ 212

    H

    .

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    P = B 1+ B 2+ M V = 13 ·H ·(B 1+ √ B 1B 2+ B 2) B1 B 2 M H

    P = 2B + M V = B ·H = r 2 ·π ·H, B M r H P = 2 ·r 2π + 2 rπ ·H = 2rπ (r + H )

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    R

    H

    2R = H,

    R

    H s

    P = B + M V = 13 BH

    13 r

    2π · H B M H r

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    P = r 2π + rπs = rπ (r + s) s

    2r = s

    R

    H s

    r

    P = B1 + B 2 + M V = 13 · H · (B 1 + √ B 1B 2 + B2) V = 1

    3 · H π

    · (R 2 + r

    · R + r 2) B1 B2

    M H R r

    P = 4r 2π V = 43 ·r 3π.

    2, 3. a H

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    P = 4 · a · h = 7200

    a2

    · h = 64800

    a = 36 h = 50

    a1 = 6 a2 = 8

    P 1 = 6a21 = 6 · 62 = 216 2 P 2 = 6a22 = 6 ·82 = 384 2

    P = 216 + 384 = 600 2 600 = 6 ·a2 a = 10

    V = 1000 3

    V = 13 BH 224 = 13 s(s −a)(s −b)(s −c) ·H s =

    a+ b+ c2 = 24

    H = 3·224√ 24·14·7·3 = 3·22484 = 8 .

    R = abc4P = 10·17·214·84 =

    858 .

    h = √ R 2 + H 2 =

    858

    2 + 8 2 = 13 , 3.

    a1 = 64 a2 = 48 b = 20

    P = a21 + a22 + 4 a 1 + a 22 H 1 H 1

    H 1 = b2 − a 1−a 22 2 = 202 − 642 −48 2 = √ 400 −64 = √ 336 H 1 P = 642 + 48 2 + 2 ·(64 + 48) ·√ 336 = 10506.

    V = H 3 (a 21 + a22 + a1a 2) H

    H = H 21 − a 1−a 22 2 = 336 − 64−482 2 = √ 272 V = √ 2723 ·(642 + 48 2 + 64 ·48) = 52072

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    a 1

    2

    bb H 1

    a 1

    2

    H 1 H 1

    H

    .

    r = 4 H = 6

    H 1

    G−H 1r 1 = H 14−r 1 ⇒

    H 1 = 24−6r 14 r 1 V = f (r 1)

    V v = B 1H 1 = πr 21H 1 = πr21(6 −

    32

    r 1) = πr 31(−32

    ) + 6 πr 21 .

    f (r 1) = −92 r 21π + 12 r 1π f (r 1) = 0 r1 = 0 r1 = 83 .

    f r1 = 83 V max = π 649 ·2 = 1289 π 3

    6 - H 1

    r 1

    H 1

    4 - r 1r

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    192π

    a > b a −b = 4 P a = 2b2π + 2bπa = 192π

    2π b2 + ba = 96

    a −b = 4b2 + ab = 96 a = 10 b = 6

    V = a2

    πb

    V = 102

    π6 = 600π.

    A D

    C B

    d H

    N

    .

    V = a22 πH ABC d2 = a2 + b2 −2ab cos α

    cos α = a2 + b2−d22ab = 122 +17 2−2522·12·17 = − 817 = cos(180 −β ) = −cos β cos β = 817 DN B

    h = bsin β = b 1 −cos2 β = 17 1 − 8172 = 15

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    M = 13 , 2 m2

    . d1 = 12 d2 = 16

    H = 247

    a = 12 b = 16 c = 20 26

    V = 768 3

    V = 4483 3

    √ 3 : 1

    cos α = √ 33 .

    a 1 = 16 a2 = 8 b = 10 V = 1344√ 3

    3

    O = 20 cm P = 24 2

    V 1 = 24π 3, V 2 = 36π 3.

    96π 2,

    V = 96π 3

    s = 5 R = 5 r = 2

    V = 104 3

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    25π

    2

    2

    V = 52π 3

    2r = 18 d = 2

    r = 3√ 38, 6 a

    a2

    V = a3 π

    2 √ 3

    3

    , P = 6a2

    π

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    A(x1y1) B (x2, y2)

    d(A, B ) = (x2 −x2)2 + ( y2 −y1)2

    C (x, y ) λ AC BC = λ

    x = x1 + λx 2

    1 + λ , y =

    y1 + λy21 + λ

    .

    λ = 1 x = x1 + x22 x = y1 + y2

    2

    AB A(x1, y1) B (x2, y2) C (x3, y3)

    ABC

    P = 12 |x1(y2 −y3) + x2(y3 −y1) + x3(y1 −y2)| .

    Ax + By + C = 0

    y = kx + n k n

    y − y1 = k(x − x1) M 1(x1, y1) k

    y − y1 = y2−y1x2−x1 (x − x1) M 1(x1, y1) M 2(x2, y2) x1 = x2. x1 = x2 xm + yn = 1 m n

    (m, n = 0)

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    tg ϕ = k2−k11+ k1 k2 k1 k2

    k1 = k2

    k2 = − 1k1 k1 = 0 M (x0, y0) Ax + By + C = 0

    d = |Ax 0 + By 0 + C |√ A2 + B 2

    C ( p, q ) r

    (x − p)2 + ( y −q )2 = r 2. O(0, 0)

    x2 + y2 = r 2.

    y = kx + n (x − p)2 +( y−q )2 = r 2

    r 2(1 + k2) = ( kp −q + n)2.

    x2 + y2 = r 2, y = kx + n r 2(1 + k2) = n 2.

    M (x0, y0) (x − p)2 +( y−q )2 = r 2

    (x − p)(x0 − p) + ( y −q )(y0 −q ) = r 2, x2 + y2 = r 2 xx 0 + yy0 = r 2

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    x

    y

    0 F 1 F 2 a

    b M

    r 1 r 2

    F 1 F 2

    2a F 1(−e, 0) F 2(+ e, 0) 0 < e < a b2x2 + a2y2 = a2b2 x

    2

    a 2 + y2b2 = 1

    b2 = a2 −e2 r 1 + r 2 = 2a a b

    = ba

    x2a 2 +

    y2b2 = 1 P = abπ

    y = kx+ n x2

    a 2 +y2b2 = 1 a

    2k2+ b2 = n 2 M (x0, y0) b2xx 0 + a2yy0 = a2b2

    xx 0a 2 +

    yy0b2 = 1

    2a

    |r 2 − r 2| = 2a x2

    a 2 − y2b2 = 1 a b

    F 1(−e, 0) F 2(+ e, 0), (0 < e < a ) e2 = a2 + b2

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    x

    y

    0 F 1 F 2

    r 1r 2

    y = ±ba x = ea > 1

    y = kx + n n2 = a2k2 −b2 M (x0, y0) b2xx 0 −a 2yy0 = a2b2 xx 0a 2 − yy0b2 = 1

    F ( p2 , 0) x = − p2 y2 = 2 px

    p ∈R \ {0}

    y = kx + n

    y2

    = 2 px

    p = 2kn

    M (x0, y0) yy0 = p(x −x0) F (0, p2 ) y = − p2

    x2 = 2 py

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    x

    y

    0 F

    M d

    r

    r = d p > 0

    x

    y

    0

    p > 0

    x

    y

    0

    p < 0

    B A(1, −2) C (5, 0) D (8, −1)

    p

    y −y1 = y2 −y1x2 −x1

    (x −x1), C (5, 0) D (8, −1)

    y −0 = −1 −0

    8 −5 (x −5),

    x + 3 y −5 = 0 p k p = −13 n A p

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    A. .

    B

    p

    n.

    y − (−2) = kn (x − 1) kn = − 1kp = 3 n y + 2 = 3( x

    −1) y

    −3x + 5 = 0

    x + 3y −5 = 0−3x + y + 5 = 0

    S (2, 1) p n B S AB xs = xA + xB2

    xB = 4 −1 = 3 yS = yA + yB2 ys = 2 + 2 = 4 B (3, 4) D (0, 0) E (3, 0) F (0, 4) ABC

    D AB E BC F AC A(x1, y1) B (x2, y2)

    C (x3, y3) 0 = x1 + x22 0 = y1 + y2

    2 3 = x2 + x3

    2 0 = y2 + y3

    2 0 = x1 + x3

    2

    4 = y1 + y32 x1 = −3 y1 = 4 x2 = 3 y2 = −4 x3 = 3 y3 = −4 P ABC = 12 |−3(−4−4)+3(3 −3)+3(4+4) | = 24

    A(−3, −3) B (−1, 3) C (11, −1)

    d(A, B ) = (−1 + 3) 2 + (3 + 3) 2 = √ 40 d(A, C ) = (11 + 3) 2 + ( −1 + 3) 2 = √ 200

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    d(B, C ) =

    (11 + 1) 2 + (

    −1

    −3)2 = √ 160

    (d(A, B ))2 + ( d(B, C ))2 = 40 + 160 = 200 = ( d(A, C ))2

    A(3, 1) B (1, −3) x

    T (xT , 0) xT = x1 + x2 + x33 yT =

    y1 + y2 + y33

    3 = 1

    2 |3(−3 −y

    3) + (y

    3 −1) + x

    3(1 + 3) | x

    T = 3+1+ x3

    3

    0 = 1−3+ y3

    3 y3 = 2 .

    3 = 12 |3(−3 −2) + (2 −1) + x3(1 + 3) | = 12 | − 14 + 4x3| = | − 7 + 2x3| = −7 + 2x3, −7 + 2x3 > 07 −2x3, −7 + 2x3 < 0

    x3

    (5, 2) (2, 2) A(3, 1)

    B (1, −3) C 1(5, 2) A(3, 1) B (1, −3) C 2(2, 2) q p : 3x +4 y

    −2 = 0

    s : x −y + 8 = 0 p s p : y = −34 x + 12

    s : y = x + 8 k p = −34 ks = 1 k q k−11+ k =

    1+ 341−34

    k = −43

    3x + 4y −2 = 0x −y + 8 = 0

    p s P (−307 , 267 ) y

    − 26

    7 =

    −43 x +

    307 4x + 3y + 6 = 0

    2x −3y + 5 = 0 3x + 2 y −7 = 0 A(2, −3)

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    6 − 512 ·8 + 2362 = r 2 1 + 512

    2 r 2 = 36 r = 6

    (x −8)2 + ( y −6)2 = 36 . (x − 3)2 + ( y − 1)2 = 8 (x −2)2 + ( y + 2) 2 = 2

    (x −3)2 + ( y −1)2 = 8(x −2)2 + ( y + 2) 2 = 2

    (x −3)2 + ( y −1)2 = 8(x −2)2 + ( y + 2) 2 = 2 ⇔ x2 −6x + 9 + y2 −2y + 1 = 8x2 −4x + 4 + y2 + 4 y + 4 = 2 ⇔

    ⇔ x2 + y2 −6x −2y = −2x2 + y2 −4x + 4y = −6 ⇔

    x2 + y2 −6x −2y = −22x + 6y = −4 ⇔⇔

    x2 + y2 −6x −2y = −2x + 3y = −2

    T 1175 , −

    95

    T 2(1, −1)

    T 2 (1 −3)(x −3) + ( −1 −1)(y −1) = 8 (1 −2)(x −2) + ( −1 + 2)( y + 2) = 2

    y = −x y = x −2 k1 = −1 k2 = 1 . k1 = − 1k2

    ϕ = π2 T 1

    x

    a + b = 36 , e = 24. a2 = b2 + e2 a2 = (36 − a)2 + 24 2 a2 = 1296 − 72a + a2 + 576 a = 26 b = √ a2 −e2 b = 10 100x2 + 676y2 = 67600

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    p p

    k p = 2

    kt = 2

    x232

    + y2

    3 = 1 a2 = 32 b

    2 = 3 n 2 = a2k2 + b2 k = 2 n2 = 9

    t1 : y = 2x + 3 t2 : y = 2x − 3

    2x2 + y2 = 32x −y = 3

    2x2 + y2 = 32x −y = −3

    P 1(−1, 1) P 2(1, −1) p : 2x − y + 4 = 0 d(P 1, p) = |2(−1)−1+4 |√ 4+1 =

    1√ 5 d(P 2, p) = |2·1+1+4 |√ 4+1 =

    7√ 5 P 1(−1, 1) p

    A(

    −5, 2)

    9x2 −4y2 = 36 y = ±32 x

    y = ±32 x + b A 2 = ±32 (−5) + b b = 2∓ 32 (−5) b1 = 192

    b2 = −112 3x −2y + 19 = 8 3x + 2y + 11 = 0 3x2 −4y2 = 72 p : 3x + 2y + 1 = 0

    p k p = −32

    kt = −32 24 −32

    2

    −18 = n 2 n = ±6 t1 : 3x + 2y − 12 = 0 t2 : 3x + 2y + 12 = 0 3x2 −4y2 = 723x + 2y = 12 3x2 −4y2 = 723x + 2y = −12

    P 1(6, −3) P 2(−6, 3) d(P 1, p) = |3·6−2·3+1 |√ 9+4 =

    13√ 13 d(P 2, p) = |−3·6+2 ·3+1√ 9+4 =

    11√ 13 P 2(−6, 3)

    m y = 52x + m

    36x2 −9y2 = 324 m

    36x2 −9y2 = 324y = 52 x + m 36x2 −9 52 x + m

    2 = 324

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    p(k2 + 2) + 2 p√ k2 + 1 − p(k2 + 2) + 2 p√ k2 + 12k2

    2

    +

    + p + p√ k2 + 1 − p − p√ k2 + 1

    k

    2

    = 16 p2

    3k4 −2k2 −1 k2 = 1 k2 = −13 , k = ±1 2x−2y− p = 0 2x+2 y− p = 0

    α = 45 k1,2 = k1±√ k2 +1

    = 11±√ 2 k1 ·k2 = −1 ⇒ α = 90

    x2 = 8 y 4x − y − 32 = 0

    y = 4x + n 25x+8 n = 0

    n = 32 4x −y−32 = 0

    M (16, 32)

    F (0, 2)

    A(8, 0)

    M (16, 32)

    P = 136 F A : x + 4 y −8 = 0 AM : 4x −y −32 = 0

    15x −8y + 16 = 0 (x −8)2 + ( y −17)2 = 289 .

    A(−2, 5) B (4, 17) AB C d(A, C ) = 2 d(B, C )

    C (8, 2)

    T (−6, 4) 4x −5y + 3 = 0 T (−4, −1)

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    A(2, −4)

    3x + 4y + 10 = 0

    ABC A(2, 6)

    BC : x −7y + 15 = 0 y −6 = 43 (x −2) y −6 = −43 (x −2)

    p : 2x−y−11 = 0 q : x−y−7 = 0 r : 3x+2 y+2 = 0 s p q

    A(2, 3)

    r

    r 3x + y −9 = 0 3x + 2 y −6 = 0 2x −3y −17 = 0.

    x + y + 2 = 0 A(1, −2) B (3, 6)

    M (−6, 4)

    x2 + 4 x + y2 −2y + 209 = 0 C (−2, 1) r = 53 .

    A(3, 4) x −y −1 = 0 r = √ 2

    (x −2)2 + ( y −3)2 = 2 (x −4)2 + ( y −5)2 = 2 A(1, 6)

    x2 + y2 + 2 x −19 = 0 t1 : y = −2x + 8 t2 : y = 12 x 112 .

    9x2 + 25 y2 = 225

    a = 5 , b = 3 = 35 F 1(−4, 0), F 2(4, 0)

    x2 +4 y2 = 36

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    y2 = 32x A B

    AB

    (x −2)2 + y2 = 64 .

    x2 + y2 = 9 y −2 = 0

    x2 = y + 3 y2 = −5(y −3)

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    a 1, a 2, a 3, . . . , a n , . . . n an n −1 an =f (a n−1) f (x)

    an n ∈N.

    an = an−1 + d a1, a 2, a 3, . . . , a n , . . . d

    n an = a1 + ( n −1) ·d

    an −an−1 = d n > 1 n S n = a1 + a n2 ·n = 2a 1 +( n−1)·d2 ·n

    an = an −k + a n + k2 n ≥ 2, k < n

    a n = an−

    1

    · q a1

    = 0 q

    = 0

    a1, a 2, a 3, . . . , a n , . . . q n

    a n = a1 · q n−1

    a na n −1

    = q n > 1 n S n = a1 1−q

    n

    1−q

    q = 1

    a2n = an−k

    ·an + k n

    ≥ 2, k < n.

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    d = 3 a1 = 1

    a25 = a1 + (25 −1) ·d = 1 + 24 ·4 = 73

    s25 = 252 (a1 + a25) = 25

    2 (1 + 73) = 25 ·37 = 925

    a1 = 10 a90 = 99 s90 = 902 (10 + 99) = 45 ·109 = 4905

    2 5

    a9 = 5 ·a2a13 = 2a 6 + 5 .

    a1 + 8 d = 5 ·(a1 + d)a1 + 12d = 2(a 1 + 5 d) + 5 ⇔ 3d = 4a 12d −5 = a1

    .

    d = 43 = a1

    a

    −d a a + d

    a −d + a + a + d = 6 (a −d)2 + a2 + ( a + d)2 = 110

    3a = 63a2 + 2 d2 = 110

    a = 2

    (2 −d)2 + 4 + (2 + d)2 = 110 , 2d2 + 12 = 110 d = ±7 −5, 2, 9

    91

    a7

    −a 2 = 20

    a3 = 9

    a1 + 6 d −a 1 −d = 20a1 + 2 d = 9 5d = 20a1 + 2 d = 9

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    d = 4 ,

    a1 = 1 . sn = n2 (2a1 + ( n

    − 1)d), n

    ∈ N

    91 = n2 (2 ·1 + ( n −1)4) 2n 2 −n −91 = 0 n = 7 n = −274 .

    n sn = 3n2 +9 n2

    a1 d

    sn = n2 (3n + 9) = n

    2 (12 + 3n − 3) = n2 (2 · 6 + 3(n − 1)) an = a1 + ( n −1)d sn = n2 (2 ·6+3( n −1))

    a1 = 6 d = 3

    b1 = −1 bn = −81 q = 3

    bn = b1q n−1 −81 = −1 · 3n−1 3n−1 = 3 4, n = 5

    23

    43

    |q | < 1

    q = q 2 a1 = a21

    a1 + a1q + a1q 2 + · · · = 23a21 + a21q 2 + a21q 4 + · · · = 43

    a11−q =

    23

    a 211−q2 =

    43

    a1 = 23 (1 −q )49 (1−q)2

    1−q2 = 43

    .

    a1 = 23 (1 −q ),13 (1 −2a + q 2) = 1 −q 2,

    a1 = 23 (1 −q ),4q 2 −2q −2 = 0 .

    q = 1 q =

    −12

    |q | < 1 q = −12 a1 = 1 −12 14 −18 , . . .

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    b1 b2 b3

    b1 + b3 = 52b22 = 100 bn = b1q n−1 b1 + b1q

    2

    = 52(b1q )2 = 100

    b1 = 2 q = 5 b2 = 10 b3 = 50

    b1 = 50 q = 15 b2 = 10 b3 = 2

    b1 = 2 q = −5 b2 = −10 b3 = 50 b1 = 50 q =

    −15 b2 =

    −10 b3 = 2

    b1

    bn = b1q n−1 b1q 2 −b1 = 3b1q −b1q 2 = 6

    b1q (q −b1) = 3b1q (1 −q ) = 6

    b1 q(1−q)b1 q(q−b1 ) =

    63 b1 = 0 , q = 1q

    q+1 = −2 ⇒ q = −23 .

    b1 = 3(−23 )2

    −1= −275

    x y z

    x + y + z = 14y + 1 −x = z −y −1y−1x = zy−1

    .

    x = 1 y = 4 z = 9 x = 9 y = 4 z = 1

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    S 8 = 96.

    n = 73.

    22

    n = 8

    a1 = 2 , a2 = 5 , a3 = 8 ,

    b1 = −1,bn = −81, q = 3 . n = 5 .

    a, aq + 8 , aq 2 a, aq + 8 ,aq 2 + 64

    49 , −209 , 1009

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    a n = 384 an−1 = 192 sn = 765

    n

    a10

    n = 8 a10 = 1536

    b7 = 5103 b1 = 7 q = 3 b7 781 b1 = 63 q = 13 .

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    n n! n n! =

    1 ·2 ·3 · · · (n −1) ·n 0! = 1 n

    nk =

    n(n−1)( n−2)···(n−k+1)k! ,k ∈Nn0 = 1

    n

    k

    (a + b)n =n0

    an +n1

    an−1b+ n2

    an−2b2 + · · ·+nn

    bn =n

    k=0

    nk

    a n−kbk

    n

    n0

    n1

    n2

    nn

    nk =

    n!k!(n−k)!

    nk =

    nn−k

    nk +

    nk+1 =

    n +1k+1

    (a + b)1

    = a + b

    (a + b)2 = a2 + 2 ab + b2

    (a + b)3 = a3 + 3 a 2b + 3ab2 + b3

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    2n .

    a = b = 1

    2n = (1 +1) n = n0 +n1 +

    n2 + · · ·+ nn

    3√ 3 + √ 2 15 .

    15

    12

    3√ 3

    3 √ 2

    12

    =15

    3 3 ·26

    = 15

    ·14

    ·13

    1·2·3 ·3 ·64 = 87360

    a2√ a + 3√ aan

    ,

    n2 = 36

    n(n−1)2 = 36 n = 9 96 a

    52

    3a−23

    6= 9·8·71·2·3 ·

    a152 −4 = 84a3√ a.

    3√ x + 1x 16 x 16k (

    3√ x)16−k 1x k = 16k x 16 −k3 ·x−k = 16k x16 −4k3

    k 16−4k3 = 0 ⇒ k = 4 x

    √ a + 1√ 3an

    103

    (n3 )

    (n

    2 ) = 103 3

    n ·(n−1)·(n−2)1·2

    ·3 = 10

    n ·(n−1)1·2 n = 12

    √ a + 1√ 3a12

    124 (√ a)8 · 1√ 3a

    4= 10·11·10·91·2·3·4

    a4 · 19a 2 = 55a2.

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    (x + y)n

    n2 − n1 = 9 n(n−1)2 −n1 = 9 n = 6

    60 +

    62 +

    64 +

    66 = 32 .

    x + 12x8

    x8 + 4 x6 + 7 x4 + 7 x2 + 358 + 74x2 + 716x4 + 1256x8 .

    √ x + 1x 15 .

    x3 + 1x318

    x

    189 .

    1x −x

    3√ x2 n x

    (

    −1)3 8

    3=

    −56

    n 5√ x + 3√ x2n

    , x2 5√ x4

    840x2 5√ x4

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    z = x + iy x, y ∈R i2 = −1 x y

    z Re (z ) = x Im (z ) = y i z 1 = x1 + iy1 z 2 = x2 + iy2

    z 1 = z 2 ⇔ x1 = x2∧y1 = y2 z 1 + z 2 = ( x1 + x2) + i(y1 + y2)

    z 1 ·z 2 = ( x1x2 −y1y2) + i (x1y2 + y1x2) z1z2 = x1 x2 + y1 y2x22 + y22 + i x2 y1−x1 y2x22 + y22 z 2 = 0

    z = x + iy z = x −iy

    x y z = x+ iy

    z = ρ(cosϕ + i sinϕ) ρ ϕ ρ = x2 + y2 tg ϕ =

    yx , x = 0

    0 x

    y . M

    z = x + iy

    i4k = 1 i4k+1 = i i4k+2 = −1 i4k+3 = −i (k ∈Z ).

    z 1 = ρ1(cosϕ1 + i sinϕ1) z 2 = ρ2(cosϕ2 + i sinϕ2)

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    z 1 ·z 2 = ρ1ρ2 [cos(ϕ1 + ϕ2) + i sin(ϕ1 + ϕ2)] z1z2 =

    ρ1ρ2 [cos(ϕ1 −ϕ2) + i sin(ϕ1 −ϕ2)] z 2 = 0

    z = ρ(cosϕ + i sin ϕ) z n = ρn (cos nϕ + i sin nϕ ), n ∈ N n√ z = n√ ρ(cos ϕ+2 kπn + i sin ϕ+2 kπn ), k = 0 , 1, 2,...,n − 1

    z = 3+4 i2−i − 5ii+3 z =

    3+ i(2−i)2

    z = 3+4 i2−i − 5ii+3 =

    3+4 i2−i ·

    2+ i2+ i − 5ii+3 · −i+3−i+3 =

    = 6+3 i+8 i+4 i2

    4−i2 + −5i2 +15 i9−i2 == 2+11 i5 − 5+15 i10 = 4+22 i−5−15i10 == − 110 + 710 i

    z = 3+ i(2−i)2 = 3+ i4−4i+ i2 =

    3+ i3−4i =

    = 3+ i3−4i · 3+4 i3+4 i =

    (3+ i)(3+4 i)9−16i2 =

    = 9+3 i+12 i+4 i2

    25 == 5+15 i25 =

    15 +

    35 i

    z = 3−2i2+ i +2−i3+ i z =

    (2+ i)(1 −2i)3−i +

    (−1+3 i)(1 −i)2+ i

    z−z1+ zz z = 1+ i

    z = 3−2i2+ i + 2−i3+ i =

    3−2i2+ i · 2−i2−i ·+2−i3+ i · 3−i3−i =

    = 6−7i+2 i24−i2 + 6−5i+ i29−i2 =

    4−7i5 + 5−5i10 =

    = 8−14i10 + 5−5i10 =

    13−19i10 = 1310 − 1910 i

    Re (z ) = 1310 Im (z ) = −1910 z = (2+ i)(1 −2i)3−i +

    (−1+3 i)(1 −i)2+ i == 2−3i−2i23−i +

    −1+4 i−3i22+ i = 4−3i3−i +

    2+4 i2+ i =

    =4

    −3i

    3−i · 3+ i3+ i +

    2+4 i2+ i ·

    2

    −i

    2−i == 12−5i−3i29−i2 + 4+6 i−4i24−i2 =

    15−5i10 + 8+6 i

    5 == 31+7 i10 =

    3110 +

    710 i

    Re (z ) = 3110 Im (z ) = 710

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    z−z1+ zz = x+ iy−(x−iy )1+( x+ iy )( x−iy ) =

    2iy1+ x2 + y2 =

    2i1+1+1 =

    23 i

    Re (z ) = 0

    Im (z ) = 23

    |z | −z = 3 −2i z = x + iy |z | = x2 + y2 x2 + y2 −x −iy = 3 −2i

    x2 + y2 −x = 3−y = −2 √ x2 + 4 = 3+ x x2 +4 = 9+ 6 x + x2 6x = −5 x = −56 . z =

    −5

    6 + 2i

    |z + 2 | < 3 3π4 < arg z ≤ 7π6 z = x + iy |z +2 | = |x + iy +2 | = (x + 2) 2 + y2 < 3, (x+2) 2+ y2 < 9

    arg z = 7π6

    x

    y

    0-5 1

    f (z ) = 2 + z + 3 z 2 f (z ) z = 3 + 2 i

    f (z ) = f (3 −2i) = 2 + 3 −2i + 3(3 −2i 2 = 20 −38i

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    a b (b = 0) a : b = ab a b

    a : b = c : d ⇒ ad = bc p% x

    p100

    p·x100 G

    p P G : P = 100 : p

    G = 100P p P