Zbirka za prijemni iz matematike

8/19/2019 Zbirka za prijemni iz matematike

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7/185

1 720

: 2, 7+2, 7 : 1, 35+

4, 2 − 1 340

·0, 4 : 2 12

= 2720

· 1027

+ 270100

· 100135

+

4210

− 4340

·4

10 · 2

5 = 12 + 2 +

168−4340

· 425 =

52 +

12540

· 425 =

52 +

12 = 3

3 34

: 7 12 − 5, 25 : 10 + 1

2 − 2

5

: 0, 2 = 15

4 : 15

2 − 21

4 : 21

2 + 1

10 : 2

10 =

154 · 2

15 − 21

4 · 2

21 + 110

· 102 =

12 − 1

2 + 12 =

12

x x

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

4 5425

: 14 75

+ 0, 8

1, 2 : 0, 375 − 0, 2 ⇔ x

1610000 :

121000 +

710

=15425

: 775

+ 810

1210 :

3751000

− 210

⇔ x1610000 ·

1000

12

+ 710

=15425

· 577 +

810

12

10 · 1000

375 − 1

5

⇔ x215

+ 710

=25 +

45

6

5 · 8

3 − 1

5

⇔

x56

=65

3 ⇔ x = 1

3.

1, 75 : 23 − 1, 75

: 712

1780

− 0, 0325 : 400 : (6, 79 : 0, 7 + 0, 3)

40 730 − 38 512

: 10, 9 +

78 − 7

30

· 1 911 · 4, 20, 08

1 7

20 : 2, 7 + 2, 7 : 1, 35 +

0, 4 : 2

1

2

·

4, 2 − 1 340

72


8/185

x

y

0

y=| x|

y = |x|

x

y

0

-1

+1

y = sgn x

y = sgn x

|x| =

x, x ≥ 0−x, x 0

0, x = 0

−1, x


9/185

(a − b)3 = a3 − 3a2b + 3ab2 − b3 a

2

− b2

= (a + b)(a − b) a3 − b3 = (a − b)(a2 + ab + b2) a3 + b3 = (a + b)(a2 − ab + b2) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc a ≥ 0 am · an = am+n a > 0 a

m

an = am−n (am)n = am·n n

√ am = a

mn

ab

n= a

n

bn

a0 = 1 (a = 0) a−n = 1an (a = 0) a, b > 0 n

√ a · b = n√ a· n√ b n ab = n√ an√ b n√ am = np√ amp = nr√ amr

m

n√

a = n

m√

a = nm√

a

133 − 1 : 0, 3−5 : 1765

1 25 · 1

7 + 2

11

− 3

4

=

133 − 10

3

−5 · 5176

75 · 25

77

− 34

=

5176

511

− 34

=

11

176

− 34

= 1634 =

24 14

3= 23 = 8

38 · 9−2 · 54 + 9 · 125 · 15−1(3 · 5)4 · 3−3 : 5 =

38 · 3−4 · 54 + 32 · 53 · 5(3 · 5)4 · 3−3 : 5 =

= 34 · 54 + 32 · 54

3 · 55 = 32 · 54(32 + 1)

3 · 55 = 32 · 54 · 10

3 · 55 = 32 · 55 · 2

3 · 55 = 6

4−14 +

1

2−32

− 43

·

4−0,25 −

2√

2− 4

3

=

(22)− 1

4 + 232

− 43

· (22)− 1

4 − 2 · 212

− 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

=


10/185

= 2−12

2

− 2−2

2= 2−1

−2−4 =

1

2 −

1

16 =

7

16

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 =

√ 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 +

√ 3 − √ 2

√ 5 + √ 3 + √ 2 ·

√ 5 +

√ 3 − √ 2

√ 5 + √ 3 − √ 2=

√ 5 +

√ 3 − √ 2

2

√ 5 + √ 32 − √ 22 =

= 5 + 3 + 2 + 2

√ 5 · √ 3 − 2√ 5 · √ 2 − 2√ 3 · √ 2

5 + 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√ 6

9 − 15 = −2√ 15 + 2√ 6

−6 =√

15 − √ 63

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√

81

4 + 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).

a2 + b2

ab − a

2

ab − b2 + b2

a2 − ab = a2 + b2

ab − a

2

b(a − b) + b2

a(a − b) =

= (a2 + b2) (a − b) − a2 · a + b2 · b

ab(a − b) = a3 + ab2 − ba2 − b3 − a3 + b3

ab(a − b) =

= ab2 − ba2ab(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 =

a2 − 2ab + b2 + 3abab

·a2 − b2

ab · a

2b2

a3 − b3 =


11/185

(a2 + ab + b2) · (a2 − b2)

a3

− b3

= (a2 + ab + b2) · (a − b) (a + b)

(a2

+ ab + b2

) (a − b) = a + b,

a = b, ab = 0, a2 + ab + b2 = 0.

23

a 3√

b

b 8√

a12+

12

√ a

a 8√

b3

: 4√ a + 4√ b , a,b > 0

23

a

3√

b

b 8√

a12+

12

√ a

a 8√

b3

:

4√

a + 4√

b =

ab13

32

ba

12

8 32 +

a

12

ab38

2 : a

14 + b

14

=a

32 b

13· 32

b32 a

128 · 32

+ a

a2b38·2

:

a14 + b

14

=

=

a

32 b

12

b32 a

94

+ a

a2b34

:

a14 + b

14

=

1

ba34

+ 1

ab34

:

a14 + b

14

=

a

14

ba +

b14

ab

:

a14 + b

14

=

1

ab.

1√

a +√

a + 1+

1√ a − √ a − 1

:

1 +

a + 1

a − 1

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

1√ a +

√ a + 1

+ 1√

a − √ a − 1

:

1 +

a + 1

a − 1

=

=

√ a − √ a + 1

a − (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


12/185

a3

−b3

a + b − aba+b

− a3 + b3

a − b + aba−b

= a3

−b3

(a+b)2−aba+b

− a3 + b3

(a−b)2+aba−b

=

= (a + b) (a3 − b3)

a2 + ab + b2 − (a − b) (a

3 + b3)

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

a2 + ab + b2 − (a − b) (a + b) (a

2 − ab + b2)a2 − ab + b2 =

= a2 − b2 − a2 − b2 = 0, a = b, a = −b


13/185

12

−8 · 16−2 + 2−3 (0, 2)6 · 50 + 13

−2 · (81−2) 14 = 3. 1, 7 · (4,5·1

23

+3,75) 713559

− 0, 5 + 13 − 512 = 1 1784 . (16−2)−2 : 16(−2)

−2: 16−2

−2= 164.

√ 3 − 95

2 − 3 85

3 − √ 32 = 125 .

1a2

+ 1b2

· a3−b3a2+b2

:

a2+b2

ab + 1

= a−b

ab , ab = 0

13√ 2+ 3√ 3 =

15

3√

4 − 3√ 6 + 3√ 9 .

23

10 2−5 5√

340

3−2 4√

220 = 1.

a−8√ ab+4ba−2 4√ ab−2√ b +

3√ b

4√ a + 4√ b

= 1.

1a2 +

1b2

a3−b3a2+b2

:

a2+b2

ab + 1

= a−bab , ab = 0.

1a+ 1

b+c

: 1a+ 1

b

− 1b(abc+a+c)

= 1.


14/185

Df ⊂ R Df ⊂ R f R× R

(∀x ∈ Df )∃1y ∈ Df y = f (x)

f : Df → Df ∃1 Df

f Df

f

f g Df = Dg

(∀x ∈ Df ) f (x) = g(x). f : Df → Df

(∀x1, x2 ∈ Df ) (x1 = x2 ⇒ f (x1) = f (x2)) .

f : Df → Df ∀y ∈ Df (∃x ∈ Df ) y = f (x). f : Df

→ Df

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

(∀x ∈ A) (f ◦ g)(x) = f (g(x)).

f : Df → Df f −1 : Df → Df

(∀x ∈ Df ) f −1 ◦ f (x) = x ∀y ∈ Df f ◦ f

−1

(y) = y.

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

x f 3(x) =

√ x2

f 4(x) = (√

x)2


15/185

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) = 2l og2 x f 2(x) = log2 x2

f 3(x) = 2log2 |x| f 4(x) = 2logx 2

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

f 1(x) = elnx

f 2(x) = x2

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

f (3) = ( 2·3−11−3 )

2 = 6.25

f

x+12x−1

= x2008 − 2x2007 + 1 f (f (2))

x x+12x−1 = 2 ⇒ x = 1

f (2) = 12008 − 2 · 12007 + 1 = 0 x x+1

2x−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+1

2

f (t) = t+12

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

) =x+12

+1

2 = x+3

4


16/185

f (x) = 5x+32x−5 g(x) =

1−x1+x

h(x) = 3−x2+x

f (x) y =5x+32x−5 ⇒ 2xy − 5y = 5x + 3 ⇒ x = 3+5y2y−5 f −1(x) = f (x)

g(x) y = 1−x1+x ⇒ y +xy = 1−x ⇒ x = 1−y1+y g−1(x) = g(x) h(x) y = 3−x2+x ⇒ 2y + xy = 3 − x ⇒ x = 3−2y1+y h−1(x) = h(x)

f (x) + 2f (1 − x) = x x ∈ R f (x) 1 − x = t ⇒ x = 1 − t f (1

−t) + 2f (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|

x−2−|x−2| = 0 x ≥ 2 0 = 0 x < 2 2(x − 2) = 0 ⇒ x = 2 Df = (−∞, 2) x ∈ {0, 1, 2, 3, 4} x ∈ {0, 1}

f (x) = 2x − 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) = f (1 − x4) = 2(1 − x4) − (1 − x4)2 = 1 − x8 f (x) =

√ 1 − x2 g(x) = sin x 6g f f −π4 +

f

g−π

4

6g

f

f −π

4

+ f

g−π

4

= 6g

f

1 − −π

4

2 +

f

sin−π

4

= 6g

f

1 − π216

+ f

−

√ 2

2

= 6g

1 − 1 − π216

+

1 − 24 = 6 sin π4 +

√ 2

2 = 7√

22

f g g (f (x)) = x2 (x) = log16 x f

−32

+ f (−1)

g−1 (x) = 16x f (x) = g−1x2

= 16

x2 = 4x

f −3

2

+ f (−1) = 4− 32 + 4−1 = 1

8 + 1

4 = 3

8


17/185

f 1(x) = sin x f 2(x) = cos x · tg x f 3(x) =

1−cos2x

2 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) = logx x

f x−1x+1

= x+2x+1 f (3) 0

f (3x + 2) = 2x − 1 f (f (x)) x−89 f (x) + 3f

1x

= x + 1 x ∈ R \ {0} f (x)

3+2x−x28x

f (x) = x(x

−1)(x

−2)(x

−3)(x+4)

log(2−x) −4, 0 f g g (f (x)) = 2x (x) = 3x + 1

f (−1) + f (−2) −6 f (x) = 1−2x2+x g(x) =

x+13−x g (f (x)) + f (g (−x))

2(7x2+8x+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+1

x+2

f (x) = x2 − 2x

f : (−∞, 1] 1−1

−→na [−1, ∞)

f −1

(x) = 1 −√

x + 1


18/185

P n(x) = anxn + an−1xn−1 + . . . + a1x + a0, an = 0,

x ∈ C n− an, an−1, . . . , a0 ∈ C

a0 = a1 = a2 = ... = an = 0 a0 = 0 ∧ n = 0 an = 1

n ≥ 1 P n(x) = anxn + an−1xn−1 + .... + a2x2 + a1x + a0,

c P n(c) = 0

x1, x2,....,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) + Rk(x).

P n(x) x − a P n( ) cC

P n(x) k c̄ P n(x)

P n(x) = anxn + an−1xn−1 + . . . + a1x + a0, an = 0

x1 + x2 + . . . + xn = −an−1anx1 · x2 + . . . + xn−1 · xn = an−2an

x1 · x2 · . . . · xn = (−1)n · a0an


19/185

n = 3 P 3(x) = a3x3 + a2x

2 + a1x + a0, a3 = 0

x1 + x2 + x3 = −a2a3x1 · x2 + x1 · x3 + x2 · x3 = a1a3

x1 · x2 · x3 = −a0a3 n = 4 P 4(x) = a4x

4 +a3x3 +a2x

2 +a1x+a0, a4 = 0

x1 + x2 + x3 + x4 = −a3a4x1 · x2 + x1 · x3 + x1 · x4 + x2 · x3 + x2 · x4 + x3 · x4 = a2a4

x1 · x2 · x3 + x1 · x2 · x4 + x1 · x3 · x4 + x2 · x3 · x4 = −a1a4x1

·x2

·x3

·x4 =

a0

a4

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

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

3x3 + 2x2

− 5x + 6

−(3x3 − 9x2 + 3x)11x2 − 8x + 6

−(11x2 − 33x + 11)25x − 5

2x2 + 3x + 11 25x − 5. x3 − 3x2 − x − 1 3x2 − 2x + 1

(x3 − 3x2 − x − 1) : (3x2 − 2x + 1)−(x3 − 2

3x2 + 1

3x)

(−73 x2 − 43 x − 1 −(−73

x2 + 149

x − 79

)

269 x

29

13

x − 79

−269

x − 29

4x5 +9x3 +19x+92 x + 1


20/185

x + 1

P (−1) = 4(−1)5

+ 9(−1)3

+ 19(−1) + 92 = 60 a P (x) = x4 + ax2 + 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 + 2x − 3

P n(1) = 3

P n(−3) = −1 P n(x) = (x2

+ 2x − 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) = (x

2 + 1) · Rn−2(x) − 2x P n(x) = (x

3 + x) · K n−3(x) + ax2 + bx + c ax2 + 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 + 2x − 3

P (x) = x4 − ax3 + bx2 + 9

Q(x) = x2 + 2x − 3 (x + 3) · (x − 1)

P (−3) = 0 ⇒ 81 + 27a + 9b + 9 = 0 ⇒ 3a + b = −10P (1) = 0

⇒ 1

−a + b + 9 = 0

⇒ −a + b =

−10

P (−3) = 0

P (1) = 0 ⇒

3a + b = −10−a + b = −10 ⇒

a = 0

b = −10


21/185

a = 0 b = −10 P (x) = x4 −10x

2

+ 9

P (x) = x4 − ax3 + bx2 + 9 = (x2 + 2x − 3) · (x2 + Ax − 3) == x4 + Ax3 + 2x3 − 6x2 + 2Ax2 − 6x − 3Ax − 9

x3 ⇒ A + 2 = a

x2

⇒ −6 + 2A = b

x ⇒ −6 − 3A = 0 ⇒ A =

−2, a = 0, b =

−10.

P (x) = (x2 + 2x − 3) · (x2 − 2x − 3) = x4 − 10x2 + 9 x1, x2, x3 125x

3 − 64 = 0 x1 · x2 ·x3 − (x1 + x2 + x3)

x1 + x2 + x3 = − a2a3 = 0 x1 · x2 · x3 = −a0a3 = 64125 x1 · x2 · x3 − (x1 + x2 + x3) = 64125

x3 + ax + b = 0 a b x1 = 1 x2 = 2

x1 + x2 + x3 = −a2a3 ⇒ 1 + 2 +x3 = 0 ⇒ x3 = −3 x1 · x2 · x3 = −6

x3 + 3x2 − 4x − 12 x1 + x2 + x3 = − a2a3 = −3 x21 + x

22 + x

23 + 2x1 · x2 + 2x1 · x3 + 2x2 · x3 = 9

x1 · x2 + x1 · x3 + x2 · x3 = a1a3 = −4 x21 + x

22 + x

23 = 17

x3 − 2x + b, b ∈ R 1 + i, i2 =

−1

1 + i x1 + x2 + x3 = −a2a3 ⇒1 + i + 1 − i + x3 = 0 ⇒ x3 = −2


22/185

a b x4 + 2x2 +

ax + b

x2

− 1 x + 6 x4 + 2x2 + 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 +5x3−12x+15 x − 2 49

a

P (x) = x4+ax3−2x2−x+3

x − 1 5 a = 4 P n(x) x +1 5

x−2 2 x2 −x−2 R

(x

) = −x

+ 4

P n(x) x6 + 1 x3 + 2

x2 + 1 R(x) = −x + 2 a, b Q(x) = x2 + x − 2

P (x) = x4 − 5x2 + ax + b a = 0; b = 4 x1, x2, x3 2x

3−x2−4 = 0 x21+x22+x23

14

x3 − 4x2 + ax + b, a, b ∈ R 1 − 2i, i2 = −1 x3 = 2, a = 9, b = −10

x1 = 2, x2 = 2, x3 =

3 −24 P 3(x) = 2x3 − 14x2 + 32x − 24 x2015 + 3x2014 + x + 3

x2 + 3x R(x) = x + 3


23/185

y = kx + n, (k, n ∈ R), (k = 0)

k

x− n

y− x = −nk k = 0 k = 0 n = 0 k = 0 n = 0

x

y

0

k > 0 n > 0

n = 0

n < 0α

y = kx + n, k > 0

ax = b x

a = 0 x = ba

a = 0

b = 0 a = 0 b = 0


24/185

x

y

0

k < 0

n > 0

n = 0

n < 0

α

y = kx + n, k < 0

x

y

0

n

y = nk = 0

n > 0

n < 0

n = 0

y = n


25/185

y = mx − 1 + 2m m y A(0, 3) m

3 = m · 0 − 1 + 2m ⇔ 4 = 2m ⇔ m = 2.

x

y

0

3

2

3

p T (6, −5) q : 2x + 3y + 5 = 0.

2x + 3y + 5 = 0 ⇔ y = −23 x − 53 . q k = − 2

3.

y + 5 = −23

(x − 6) ⇔ y = −23

x − 1. M (−5, 4)

P = 5

M (

−5, 4) y

−4 = k(x +

5) ⇔ y = kx + 5k + 4 m n xm +

yn = 1,

P = 12|mn|.

M

−5m

+ 4n

= 1 5


26/185

x

y

0

M (-5,4)

-5

4

n

m

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 + 144

m2 = 25 ⇔ 144

9 = m2 ⇔ m2 = 16 ⇔

m = ±4

x−12 +

3x−14 =

2x−43 +

x+16

N ZS (2, 3, 4, 6) 6(x−1)+3(3x−1) = 4(2x−4)+2(x + 1) ⇔ 15x−9 = 10x−14 ⇔5x = −5 ⇔ x = −1

1x2+2x+1 +

2x+2x2+x3 =

52x+2x2

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


27/185

x − 1+ 34x4

+ 5− 2

3x

4 =

3−x2

3

N ZS (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 ⇔ 2x

2+8x2−4 =

2 ⇔ 2x2+8x2−4 − 2x

2−4x2−4 = 0 ⇔ 16x2−4 = 0.

|x + 2| − |2x − 1| = 1

|x + 2| =

x + 2, x ≥ −2−(x + 2), x 2 ∧ x = 0 ⇔ x ∈ ∅


28/185

{−1, 1}. 2|x + 1| − |x − 2| − 3 = 0

|x + 1| =

x + 1, x ≥ −1−(x + 1), x 0 ⇔ 6x − 9 − 1 − 3x + 14 − 8x > 0 ⇔−5x + 4 > 0 ⇔ x < 4

5.

2|x − 2| − |x + 3| + 2 > 0

x ∈ (−∞, −3) x − 2 < 0 x + 3 < 0

−2(x − 2) + (x + 3) + 2 > 0 ⇔ −x + 9 > 0 ⇔ x < 9 (−∞, −3) ∩ (−∞, 9) = (−∞, −3)

x ∈ [−3, 2] x−2 < 0 x+3 > 0 −2(x−2)−(x+3)+2 > 0 ⇔ x 2 x−

2 > 0 x + 3 > 0 2(x − 2) − (x + 3) + 2 > 0 ⇔ x − 5 > 0 ⇔ x > 5 (2, +∞) ∩ (5, +∞) = (5, +∞)

(−∞, −3) ∪[−3, 1) ∪ (5, +∞) = (−∞, 1) ∪ (5, +∞).


29/185

x + 7y − 12 = 0 2x − y + 6 = 0 A(8, −4) 2x + 5y − 4 = 0 x − 3y = 0 AB A(−1, 7) B(6, 8) π

4

C (3, −1). kx + (k + 1)y − p = 0 k p M (2, 1)

x + 2y − 4 = 0

5x−23 − 13x+1

7 = x−5

2 + x x = 1.

x+34 − 2x−1

2 = 7x−1

3 + 1 7

12. x = 4.

1x3−27 +

2x−3 =

2x+1x2+3x+9

. x = −2.

x1+ x

1−x= x2. 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, ∞).


30/185


31/185

a > 0 a 0

x

y

0 x1 x2 x

y

0

x1 x2

D = 0

x

y

0 x1= x2

x

y

0

x1= x2

D


32/185

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

m = −2 4x2 + mx + m2 − 15 = 0 m

D = m2 − 4 · 4 · (m2 − 15) = −15m2 + 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) x

−3 4 y = (x−4)(x+3) −3 4 x2

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.


33/185

x

y

0-3 4

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 ⇔ 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)


34/185

(−∞,−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)

x ∈ (−∞, −2] ∪ (−1, 0) ∪ [1, 3] m

x2 − 2mx + m2 − 1 = 0 [−2, 4]

x1,2 = 2m

±

√ 4m2

−4·1·(m2

−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−4

2(m−1) = 1 m = 2

y = 2x2 + bx − 3 b

x = 5

4

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) 2|x + 1|


35/185

ax4 + bx2 + c = 0 a(x2)2 + bx2 + c = 0 x2 x2 = t at2 + bt + c = 0

ax4 + bx3 +cx2 + bx + a = 0 x2

x + 1x = t

axn±b = 0 a, b n = 3 ax3±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 − (m2 + n2)x2 + m2n2 = 0

x2 = t t2

−(m2 + n2)t +

m2n2 = 0 t1,2 = +(m2+n2)±√ (m2+n2)2−4·1·m2n2

2 =

+(m2+n2)±√ m4−2m2n2+n42

= (m2+n2)±

√ (m2−n2)2

2 = (m

2+n2)±|m2−n2|2

t1 = x2 = m2 t2 = x

2 = n2 x1 = m x2 = −m x3 = n x4 = −n.


36/185

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

x + 6

x2 = 0 ⇔

6(x2 + 1x2 ) − 35(x + 1x) + 6 = 0 x + 1x = t x2 + 2 + 1

x2 = t2 x2 + 1

x2 = t2 − 2

6t2 − 35t + 50 = 0. t = 10

3 ∨ t = 5

2.

x + 1x = 10

3 x + 1x =

52

. x2 −103

x + 1 = 0 x1 = 3 x2 = 1

3.

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.

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. 4x4 + 2x3 + 6x2 + 3x + 9 = 0 x1,2 =

1±i√ 52

x3,4 = −3±i√ 15

4

x4 + 1 = 0 x1,2 =√

2(−1±i)2 x3,4 =

√ 2(1±i)

2

(2x − 4) 12 − (x + 5) 12 = 1 x = 20. √ x + √ x + 9 = √ x + 1 + √ x + 4 x = 0.

√ x + 4 < x − 2. x ∈ (2, 5).


37/185


38/185

√ x − 9 − √ x − 18 = 1

√ 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+4x−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 > x2 − 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)


39/185

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)

√ x+2

2 = 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

√ 2x2 − 7x + 3 ≥ 0 x ∈ (−∞, 12 ] ∪ [3, ∞)

√ x2 − x − 12 < x. x ∈ [3, ∞)


40/185

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

1

x

y

0

y = a

x

0 < a < 1

1

ax1 = ax2 ⇔ x1 = x2 af (x) > b b ≤ 0 af (x) f (x) b > 0 f (x) > loga b a > 1 f (x) < loga b 0 < a


41/185

x

y

0

y = 2 x

1

y = ( ) x12

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

x


42/185

x

y

0

y = 3 x-1

-1

x

y

0

y = |3 x-1|

1

2x−

1

= 45

9−3x

= 1

27x+3

3

x−12 − 2

x+1

3 =2x−23 + 3

x−32

0, 125 ·42x−3 =√

28

−x 10 ·2x−4x = 16 2 ·3x+1 −

5 · 9x−2 = 81

2x−1 = 45 ⇔ 2x−1 = (22)5 ⇔ 2x−1 = 210 ⇔ x − 1 = 10 ⇔ x = 11 9−3x =

1

27

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

x−12 − 2x+13 = 2x−23 + 3 x−32 ⇔ 3 x−12 − 3x−32 = 2x−23 + 2 x+13 ⇔ 3 x−32 (3 − 1) =

2x−23 (2 + 1) 2

· 3

x−32 = 3

· 2

x−23

⇔ 3

x−32

· 3−1 =

2x−33 · 2−1 ⇔ 3 x−52 = 2x−53 ⇔ 3 12x−5 = 2 13x−5 ⇔ √ 33√ 2x−5 = 1 ⇔ x − 5 =

0 ⇔ x = 5 0, 125 · 42x−3 =

√ 2

8

−x⇔ 2−3 · 24x−6 =

2−

52

−x⇔ 24x−9 = 2 52x ⇔

4x − 9 = 52 x ⇔ x = 6


43/185

10 · 2x − 4x = 16 ⇔ 10 · 2x − x2x = 16 2x = t 10t − t

2

= 16

t1 = 8 t2 = 2 2x = 23 2x = 21

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 = 34 3x = 815

x1 = 4 x2 = 4 − log 5log 3

5x − 3x+1 > 2 (5x−1 − 3x−2) 143x−2 ≤

12|x+1|

3x 3x > 0 x

5x − 3x+1 > 2 (5x−1 − 3x−2) ⇔ 53x − 3 > 2 · 15 · 53x − 29 ⇔ 35 53x > 3 − 29 ⇔35

53

x> 25

9 ⇔ 5

3

x>

53

3

53

> 1 x > 3

14

3x−2 ≤ 12

|x+1| ⇔ 12

23x−2 ≤ 12

|x+1| ⇔ 12

6x−4 ≤ 12

|x+1|

12

< 1 6x − 4 ≥|x + 1| ⇔ [(6x − 4 ≥ x + 1 ∧ x + 1 ≥ 0) ∨ (6x − 4 ≤ −(x + 1) ∧ x + 1 < 0)]⇔ [(5x ≥ 5 ∧ x ≥ −1) ∨ (7x ≥ 3 ∧ x


44/185

y = 2x − 2 y = 14x + 1

x

y

0

y = 2 x-2

-1

-2

x

y

0

y = ( ) x

+11

4

1

2

x

−7

√ 32x+5

= 0, 25 · 128x+17

x−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| ≤ 14

2x−3 x ∈ (−∞, 3]


45/185

x > 0 a (a > 0, a = 1) loga x c a x c = loga x ⇔ ac = x aloga x

loga 1 = 0

loga a = 1

loga

(b·

c) = loga

b + loga

c

logabc = loga b − loga c

loga bx = x · loga b

loga b = logc blogc a

loga b = 1logb a

logak b = 1k loga b

log10 x = log x e (e = 2.71828...) loge x = ln x


46/185

x

y

0 1

y = loga x

a > 1

y = loga x a > 1

x

y

0 1

y = loga x

0 < a < 1

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


47/185

log 12 x = 3 ⇔ x = 123 = 18 x1+log3 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 log x(x+9) = 1 ⇔ log10 x(x+9) = log10 10 ⇔ x(x+9) = 10∧x(x+9) > 0 x2+9x+10 = 0 x1 = 1 x2 = −10 x(x+9) x = 1

logan b = 1n loga b log3 x + log9 x +

log81 x = 7 ⇔ log3 x + 12 log3 x + 14 log3 x = 7 ⇔ 74 log3 x = 7 ⇔ log3 x = 4 ⇔x = 34 = 81 ∧ x > 0 log4(x+12)·logx 2 = 1 x + 12 > 0 ∧ x > 0 ∧ x = 1 log4(x + 12) · logx 2 = 1 log4(x + 12) = log2 x ⇔ 12 log2(x + 12) = log2 x log2(x + 12) = log2 x

2 x + 12 = x2 x1 = 4

x2 = −3 x = 4 log49 x

2 + log7(x − 1) = log7(log√ 3 3) x = 0 ∧x − 1 > 0 log72 x2 + log7(x − 1) = log7(log31/2 3) ⇔12 log7 x

2 +log7(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 ⇔log x = log400

log 8000 = 2+2log2

3+3log2 = 2

3 ⇔ x = 10 23

log(2x + 3) ≤ 1

log13

3x

−1

x+2 > 1

log(x +1) < log(2x − 1)

log(2x + 3) ≤ 1 ⇔ log(2x + 3) ≤ log 10 ⇔ 2x + 3 ≤ 10 ∧ 2x + 3 > 0 ⇔x ≤ 7

2 ∧ x > −3

2 ⇔ x ∈ (−3

2, 7

2]


48/185

log 13

3x−1x+2

> 1 ⇔ 0 < 3x−1x+2

< 13 ⇔ 0 < 3x−1

x+2 ∧ 3x−1

x+2 < 1

3

x ∈ (−∞, −2) ∪ (13 , +∞) = A

8x−5x+2

0 ∧ 2x − 1 > 0) ⇔3 < x ∧ x > −2 ∧ x > 12

⇔ x > 3

log2(x − 4) = 3 x = 12

log x

log x−log3 = 2 x = 9,

1 +

log2 x − 1 = log x x = 10, 2x2 = (2x + 5) logx 4 · log8 x x = 53

logx(x + 2) > 2 x ∈ ∅ log8(x

2 − 4x + 3) < 1 x ∈ (−1, 1) ∪ (3, 5) log

√ 2 − x − log √ 4 − x2 + 3 log √ 2 + x


49/185

α ABC sin α = a

c =

cos α = bc =

tg α = ab

=

ctg α = ba =

b

a

c

α

∠ pOq k1(O, r1) k2(O, r2)

l1r1

= l2r2

lr = 1, l

r

1

π6

π4

π3

π2

π 3π2

2π

30 45 60 90 180 270 360


50/185

p

q

O r 1

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

sin x = xM cos x = yM tg x = xM yM

, yM = 0 ctg x =yM xM

, xM = 0


51/185

Oq

2π

xM

yM

yM

xM

π

y = sin x, x ∈ R y = cos x, x ∈ R 2π y = tg x, x = π

2 + kπ, k ∈ Z y = ctg x, x =

kπ, k ∈ Z π

f (x) \ x 0 π6 π4 π3 π2 π 3π2 2πsin x 0 1

2

√ 2

2

√ 3

2 1 0 −1 0

cos x 1√

32

√ 2

212 0 −1 0 1

tg x 0√

33

1√

3 − 0 − 0ctg x − √ 3 1

√ 3

3 0 − 0 −

x

y

0 x

y

0 x

y

0

sin x cos x tg x, ctg x

++

- -

+

+-

- +

+

-

-

sin(2kπ + α) = sin α, k ∈ Zcos(2kπ + α) = cos α, k ∈ Zsin(π

2 − α) = cos α,

cos(π2 − α) = sin α,sin(π − α) = sin α,


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cos(π − α) = − cos α,sin(

π

2 + α) = cos α,cos(π2

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

2 − α) = − sin α,

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

2 − α) = sin α,

0

−α sin(0

−α) = − sin α cos(0−α) = cos α

sin2 α + cos2 α = 1

tg α = sinαcosα , α = π2 + kπ, k ∈ Z ctg α = cosαsinα , α = kπ, k ∈ Z tg α · ctg α = 1, k = kπ2 , k ∈ Z

sin2 α = sin2 α

1 = sin2 α

sin2 α+cos2 α = tg

2 αtg2 α+1

, α = π2 + kπ, k ∈ Z

cos2 α = cos2 α

1 = cos2 α

sin2 α+cos2 α = 1tg2 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 β

1−tg α·tg β α , β , α + β = π2 + kπ, k ∈ Z


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

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 α

1−tg2 α , α, 2α = π2 + kπ, k ∈ Z,

ctg2α = ctg2−1

2ctg α , α, 2α = kπ, k ∈ Z.

2cos2 α2

= 1 + cos α cos α2

= ±

1+cosα2

2sin2 α2

= 1 − cos α sin α2

= ±

1−cosα2

tg α2 = ±

1−cosα1+cosα

, α = π + 2kπ, k ∈ Z,

ctg α2 = ±

1+cosα1−cosα , α = 2kπ, k ∈ Z.

sin α + sin β = 2sin α+β2 cos α−β

2

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+β

2 sin α−β

2

tg α ± tg β = sin(α±β)cosα cosβ , α, β = π2 + kπ, k ∈ Z,

ctg α ± ctg β = sin(β±α)sinα sinβ , α, β = kπ, k ∈ Z,


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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 )]

x

y

0

1

-1

2

-

2

3

22

y = sin x

y = sin x

x

y

0 1

-1

2

2

y = arcsin x

y = arcsin x


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x

y

0

1

-1

2

-

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 x

2

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 −1 −√

32

−√

22

−12 0

12

√ 2

2

√ 3

2 1

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

63π4

2π3

π2

π3

π4

π6

0

f (x)

\x

→ −∞ −√

3

−1

−

√ 3

3 0

√ 3

3 1

√ 3

→ ∞arctg x → −π2

+ 0 −π3

−π4

−π6

0 π6

π4

π3

→ π2

+ 0arcctg x → π − 0 5π6 3π4 2π3 π2 π3 π4 π6 → 0 + 0

sin x = a |a| ≤ 1 ⇔ x = arcsin a + 2kπ ∨ x = π − arcsin a + 2kπ k ∈ Z. cos x = a, |a| ≤ 1 ⇔ |a| ≤ 1 ⇔ x = arccos a + 2kπ ∨ x = −arccos a +

2kπ,k ∈ Z. tg x = a, a ∈ R ⇔ x = arctg a + kπ, k ∈ Z. ctg x = a, a ∈ R ⇔ x = arcctg a + kπ, k ∈ Z. sin ax ± sin bx = 0 cos ax ± cos bx tg ax ± tg bx = 0

ctg ax ± ctg bx = 0

a sin x + b cos x = 0, a, b = 0


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x = π2

+ kπ,k ∈ Z. cos x

a sin x + b cos x = c, a,b, c = 0, |c| a (1 ≤ a ≤ 1) ⇔ arcsin a+2kπ < x < π−arcsin a+2kπ, k ∈ Z. sin x < a (1 ≤ a ≤ 1) ⇔ π − arcsin a < x a (1 ≤ a ≤ 1) ⇔ − arccos a + 2kπ < x


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1+ctg 2x·ctg xtg x+ctg x

= 12 ctg x sinα+cosα

sinα−cosα− 1+2cos2 α

cos2 α(tg2 α−1) =

21+tg α , sin4

α+2sinα cosα−cos4

αtg 2α−1 = cos 2α cosα−sinα−cos3α+sin3α2(sin 2α+2cos2 α−1) = sin α

8cos3 x−2sin3 x+cosx2cosx−sin3 x = −32 tg x = 2

1+ctg 2x·ctg xtg x+ctg x

= 1+ cos 2x

sin 2x· cosxsinx

sinxcosx

+ cosxsinx

=sinx sin 2x+cos2x cosx

sin 2x sinxsin2 x+cos2 xsinx cosx

=cos(2x−x)sin 2x sinx

1sinx cosx

cos2 x sinxsin2x sinx

= cos2 x

2sin x cosx =

12 · cosx

sinx= 1

2 ctg x

sinα+cosαsinα−cosα− 1+2cos

2 αcos2 α(tg2 α−1) =

sinα+cosαsinα−cosα− 1+2cos

2 α

cos2 αsin2 αcos2 α

−1 = sinα+cosαsinα−cosα− 1+2cos

2 αsin2 α−cos2 α =

(sin α+cosα)2

sin2 α−cos2 α − 1+2cos2 α

sin2 α−cos2 α = sin2 α+2sinα cosα+cos2 α−1−2cos2 α

sin2 α−cos2 α =2sinα cosα−2cos2 α

sin2 α

−cos2 α

= 2cosα(sinα−cosα)(sinα−

cosα)(sinα+cosα) = 2cosαsinα+cosα =

2sinα

cosα

+1 = 2tg α+1

sin4 α+2sinα cosα−cos4 αtg 2α−1 =

(sin2 α)2−(cos2 α)2+sin2αtg 2α−1 =

(sin2 α+cos2 α)·(sin2 α−cos2 α)+sin2αtg 2α−1 =

(sin2 α−cos2 α)+sin2αtg 2α−1 =

− cos2α+sin 2αtg 2α−1 =

sin2α−cos2αsin 2αcos 2α

−1 = sin2α−cos2αsin 2α−cos 2α

cos 2α

= cos 2α

cosα−sin α−cos3α+sin3α2(sin 2α+2cos2 α−1) =

(cosα−cos3α)+(sin3α−sinα)2(sin 2α+cos2α)

= 2sin2α sinα+2cos2α sinα2(sin 2α+cos 2α)

=2sin α(sin2α+cos2α)

2(sin 2α+cos2α) = sin α

tg x = 2 sin x = 2cos x sin x 8cos

3 x−2·8cos3 x+cosx2cos x−8cos3 x =

cosx−8cos3 x2(cos x−4cos3 x) =

cosx(1−8cos2 x)2cosx(1−4cos

2

x)

= 1−8cos2 x

2(1−4cos2

x)

= sin2 x+cos2 x−8cos2 x

2sin2

x+2cos2

x−8cos2

x

= sin2 x−7cos2 x

2sin2

x−6cos2

x

cos2 x tg2 x−7

2 tg2 x−6 = 4−78−6 = −32 .

x

tg x = 34 , π < x ≤ 3π2

sin x = ± tg x√ 1+tg2 x

sin x = ± 34√ 1+ 9

16

= ± 35 π < x ≤ 3π

2 sin x = − 3

5

cos x = −45 , ctg x = 43

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

2

1cosx

=cos x + sin x cos2x + sin2 x = cos x

cos x cos π5 + sin x sin π5 =

√ 3

2

−π4

, 9π4

sin x2

+ cos x = 1.


60/185

sin(x + π

4 ) = 1

x + π4

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

2

t = π2 + 2kπ ∨ t = π − π2 + 2kπ. π − π2 = π2 t = π

2+2kπ, k ∈ Z. x+π

4 = π

2+2kπ, k ∈

Z x = π4 + 2kπ, k ∈ Z. sin x + cos x = sin x cos x + 1 ⇔ sin x − sin x cos x + cos x − 1 = 0 ⇔sin x(1 − cos x) − (1 − cos x) = 0 ⇔ (sin x − 1)(1 − cos x) = 0, sin x − 1 = 0 1 − cos x = 0 sin x = 1 ∨ cos x = 1 ⇔x = π

2 + 2kπ

∨x = 2kπ, k

∈Z.

sin2 x + sin2 2x + sin2 3x = 32 ⇔ 1−cos2x2 + 1−cos4x2 + 1−cos6x2 = 32 ⇔32 − 1

2(cos 2x + cos 4x + cos 6x) = 3

2 ⇔ cos2x + cos 4x + cos 6x = 0 ⇔

(cos 6x + cos 2x) + cos6x = 0 32 ⇔ 2cos 6x+2x2 cos 6x−2x2 + cos4x = 0 ⇔2cos4x cos2x +cos4x = 0 ⇔ cos 4x(2cos2x + 1) = 0 cos4x = 0 cos2x = − 12 4x = π2 + 2kπ 2x = ± 2π3 + 2kπ k ∈ Z x = π

8 + kπ

2 x = π

3 + lπ x = π

3 + mπ

k , l , m ∈ Z cos x = 0 1

cosx = cos x + sin x

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

sinx

(sinx

− cosx

) =0 ⇔ sin x = 0 ∨ sin x = cos x ⇔ x = kπ ∨ x = π4 + lπ, k, l ∈ Z cos2x +sin2 x = cos x ⇔ cos2 x− sin2 x +sin2 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 cos π5 + sin x sin

π5 =

√ 3

2 ⇔ cos(x− π5 ) =

√ 3

2 ⇔ x − π

5 = ±π6 + 2kπ, k ∈Z, x = π

5 ± π

6 + 2kπ, k ∈ Z ⇔ x = π

30 + 2kπ, k ∈ Z∨x = 11π

30 + 2lπ, l ∈ Z

π30

, 61π30

11π30

sin x2

+ cos x = 1 ⇔ sin x2

+

1 − 2sin2 x2

= 1 ⇔ sin x

2 − 2sin2 x

2 = 0 ⇔

sin x2

1 − 2sin x2

= 0 ⇔ sin x2 = 0 ∨ sin x2 = 12

x2

= kπ, k

∈ Z

⇔ x = 2kπ, k

∈ Z

x2

= π6

+ 2lπ, l

∈ Z

∨ x

2 =

π − π6 + 2mπ, m ∈ Z x = π3 + 4lπ, l ∈ Z ∨ x = 5π3 + 4mπ, m ∈ Z. 2cos x + 1 ≥ 0 √ 3 tg x


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2cos x + 1 ≥ 0 ⇔ cos x ≥ −12

cos x = −1

2

x = ±2π3

+2kπ, 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

√ 3 tg x


62/185

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

4

x = π4 −3π4

+ 2kπ < x < π4

+ 2kπ, k ∈ Z

x

y

0

-1

2

-

23 2

y = sin x

y = cos x

4

3

4 2

sin −√ 22 cos −

√ 2

2

sin75 + sin15 √

62


63/185

2·ctg α1+ctg2 α

= sin 2α

sin2α1+cosα ·

cosα1+cos 2α = tg

x2

1−cosxsinx =

sinx1+cosx = tg

x2

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

2 cos α

2 tg α

2 cos α = 119

169 sin α

2 = ± 5

13, cos α

2 =

±1213

, tg α2 = 512

sin3x = −√

22

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

12 + kπ

2

tg 4x = ctg 6x x = π

20

+ kπ

20

, k

∈Z

√ 3sin x + cos x = √ 2 x = π12

+ 2kπ ∨ x = 7π12

+ 2lπ, k, l ∈ Z

sin x ≤ cos 3x −π4

+ 2kπ ≤ x ≤ π8

+ 2kπ, k ∈ Z 5π8

+ 2kπ ≤x ≤ 3π

4 + 2kπ, k ∈ Z 9π8 + 2kπ ≤ x ≤ 13π8 + 2kπ, k ∈ Z ctg x ≥ ctg π

11 kπ < x ≤ π

11 + kπ, k ∈ Z

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


64/185

A B

C

ab

c

hc

a, b, c

α, β, γ

α1, β 1, γ 1

ha, hb, hc

s

r

R

O = a + b + c = 2sP = a·ha

2 = b·hb

2 = c·hc

2

P =

s(s − a)(s − b)(s − c)P = a·b·c4R = r · sP = 1

2 · a · b · sin γ = 1

2 · a · c · sin β = 1

2 · b · c · sin α

a = b = c

O = 3a, P = a2

√ 3

4 , h = a√

3

2 , r = a√

3

6 , R = 2r = a√

3

3 . c a b

c2 = a2 + b2 P = ab

2 , R = c

2, a2 = c · p h2c = p · q b2 = c · q


65/185


66/185

A B

C D

d 2

d 1

.

ha

P = a

·h ha a

P = a · b O = 2a + 2b.

A B

C D

. .

..

d

d

a

b

P = a2, P = d2

2 O = 4a

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