Zbirka Zadataka Za Prijemni Iz Matematike VA

8/16/2019 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


7/213

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


8/213

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.


9/213

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.


10/213

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 =


13/213

=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


20/213

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


21/213

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


22/213

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


23/213

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


24/213

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


25/213

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


26/213

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


27/213

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


28/213

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


29/213

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}.


30/213

|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}.


31/213


32/213

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 , ∞)


33/213

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


34/213

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


35/213

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


36/213

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)


37/213

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)


38/213

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 =


39/213

= (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 .


40/213

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


41/213

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


42/213

√ 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)


43/213

√ 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 .


44/213

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


45/213

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


46/213

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


47/213

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, ∞).


48/213

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]


49/213


50/213

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


51/213

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


52/213

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)


53/213

log12

log8 x2−2xx−3

< 0

x ∈ (3, 4)∪(6, ∞)


54/213

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


55/213

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


56/213

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

++- -

+

+-- +

+- -


57/213

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


58/213

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 )]


59/213

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


60/213

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


61/213

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


62/213

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 .


63/213

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 ,


64/213

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,


65/213

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


66/213

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


67/213

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


68/213

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


69/213

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


70/213

−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 .


71/213

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


72/213


73/213

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


74/213


75/213

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


76/213

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


77/213

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


78/213

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 .


79/213

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


80/213

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


81/213

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


82/213

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


83/213

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


84/213

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

.


85/213

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 )


86/213

R

H

2R = H,

R

H s

P = B + M V = 13 BH

13 r

2π · H B M H r


87/213

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


88/213

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


89/213

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


90/213

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


91/213


92/213


93/213

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


94/213

25π

2

4π

2

V = 52π 3

2r = 18 d = 2

r = 3√ 38, 6 a

a2

V = a3 π

2 √ 3

3

, P = 6a2

π


95/213

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)


96/213

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


97/213

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


98/213

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


99/213

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


100/213

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


101/213

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)


102/213


103/213

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


104/213


105/213

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


106/213


107/213

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)


108/213

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


109/213


110/213

y2 = 32x A B

AB

(x −2)2 + y2 = 64 .

x2 + y2 = 9 y −2 = 0

x2 = y + 3 y2 = −5(y −3)


111/213

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.


112/213

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


113/213

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 , . . .


114/213

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


115/213

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


116/213

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 .


117/213

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


118/213

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.


119/213

(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


120/213

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)


121/213

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


122/213

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


123/213


124/213

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

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