2008.03.09 Държавен зрелостен изпит по МАТЕМАТИКА

Page 1

3

2008 . –

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28. 26.

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1.

20.

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1.

3+ 5 )0

)– 2 5 x2 − 5x + 4 x2 − 4x

2. )

x−5 x −1

x1 3. 1 1 + x1 x2 ) −

)

x −1 x−4

)

x −1 x

)

57 2

: 4 5

)

2sin α.sin ( 90° − α )

cos α − cos (180° − α ) ) sin α

)

5 4

α ∈ [ 0°;90° )

: ) 2sin α

-

) log 4 64

)1

?

2 sin 45°

)

( −5)

2

) 5log5 6

x4 − 8x2 − 9 = 0 ?

6.

) ±3

±3

) ±1

–1

)1

–9

5 − x = x − 5?

7. )5

)5

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4

) 10

8. ) ∀x ≤ −2

10.

3 2

:

4 − 5x −4 x

5.

)9

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2 x2 + 5x − 4 = 0 ,

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) cos α

?

x≠0

x2

5 4

4.

3− 5

) 2 5

x≠4 )

2

) ∀x ≤ ±2

4 x 2 − 16 ≤ 0 ? ) ∀x ∈ ( −∞; −2] ∪ [ 2; +∞ )

-

y = x 2 + 6 x + 10 ? ) 10

) –3

-

) –2

) –6 3 4

16 25 )8 5

11. ) 60

2

1 2

)∅

8

) ∀x ∈ [ −2; 2]

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α ∈ [ 0°;90°] ? )2

) 40

) 10

y = 2 cos 3α

:

1


12. )

1 3

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13. m. )7m

ABC

14. B;D, ) 36

a b OC = 4 cm, AB = 9 cm ) 4,5

1 18

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1 36

CD = 13

AB = 6 m ?

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HXOY

A;C x?

OA = CD = x cm. )6

) 10

) 2 cm

) 18 cm 2

2

) 36 cm 2

81 ° 10 3 cm. ? ) 30 cm

) 54 cm 2 , AH = 4 cm

BH = 9

) 78 cm 2

39 ° ,

) 20 cm

) 40 cm -

18. cm 2 . ) 8 cm

?

ABC( HACB = 90°) , CH J ABC ? ) 13 cm 2

17.

) 60 cm

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15.   6 cm 2 . ) 9 cm 2 16. cm. ) 39 cm 2

22 m

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12

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19. ) 60

20. ABCD   BP : PD = 4 : 3 AC = 25 cm. P, AB AD? )4 3 ) 5 10 6

2

) 16 cm

) 18 cm BC = 3 cm,

) 90

HBAC ? ) 30

HBAD

) 20

15

BD

) 16

9

2


21.

25.

α

21.

sin α =

cos α + sin α. tg α ?

x1

22.

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!

4 , 5

x 2 + kx + 1 = 0 ,

x2

k

x1 + x2 ≤ 2 ? 2

ABCD

23.

2

BAD = 30 ° , AD = 2 3 cm CD ? 13 cm, 14 cm

24.

BD = 4 cm.

15cm. ?

25.

?

26.

28.

.

! 26.

21 − ( x2 − 4 x ) = 6 . x − 4 x + 10

27.

R

2

AB = 2 R . ABCD, 2 R ⎡⎣1 + cos α + cos β − cos ( α + β ) ⎤⎦ ,

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α = HBAD α = β, 28.

2

β = HABC .

α

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9.

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ax 2 + bx + c = 0

−b ± b 2 − 4ac ax 2 + bx + c = a ( x − x1 )( x − x2 ) 2a b c x1 + x2 = − x1 x2 = a a

x1,2 =

(−

y = ax 2 + bx + c , a ≠ 0

b D ;− ) 2a 4a

. 2k

2 k +1

a 2k = a

m n m a =an log a b = x ⇔ a x = b

nk

a 2 k +1 = a ;

a mk = n a m

log a a = x x

k∈

a = nk a ; a loga b = b ;

k-

: k-

n

b > 0, a > 0, a ≠ 1

Vnk = n. ( n − 1) ... ( n − k + 1)

: Cnk =

Vn k n. ( n − 1) ... ( n − k + 1) = Pk 1.2.3...(k − 1)k

P ( A) =

:

an = a1 + ( n − 1) d

:

an = a1.q n −1

p ⎞ ⎛ : K n = K .q = K . ⎜ 1 + ⎟ ⎝ 100 ⎠ n

n, m, k ∈

Pn = 1.2.3... ( n − 1) n = n !

:

n n

a > 0, n ≥ 2, k ≥ 2

n k

0 ≤ P( A) ≤ 1

2a + ( n − 1) d a1 + an ⋅n = 1 ⋅n 2 2 a q − a1 qn −1 Sn = n = a1 ⋅ q −1 q −1

Sn =

n


1 1 ab = chc a 2 = a1c b 2 = b1c 2 2 + − c a b a b a b 2 hc = a1.b1 r= sin α = cos α = tgα = cotgα = 2 c c b a 2 2 2 2 2 2 b = a + c − 2ac cos β : a = b + c − 2bc cos α a b c c 2 = a 2 + b 2 − 2ab cos γ = = = 2R sin α sin β sin γ 1 1 2 : ma = ( 2b 2 + 2c 2 − a 2 ) mb 2 = ( 2a 2 + 2c 2 − b 2 ) 4 4 1 mc 2 = ( 2a 2 + 2b 2 − c 2 ) 4 a n   : = lc2 = ab − nm b m : c 2 = a 2 + b2

:

:

1 S = chc 2

S=

1 ab sin γ 2 abc S = pr S= 4R S = ab sin α S = aha 1 : S = d1d 2 sin ϕ 2 : S = pr

α0

00

α rad

0

sin α

0

cos α

1

tg α

0

cotg α

p ( p − a )( p − b )( p − c )

S=

S=

300

450

600

900

π

π

π

π

6 1 2

4 2 2 2 2

3 3 2 1 2

2

1

3

1

3 3

0

3 2 3 3 3

1 0


−α − sin α cos α − tgα −cotgα

sin cos tg cotg

900 − α cos α sin α cotgα tgα

sin (α ± β ) = sin α cos β ± cos α sin β

tgα ± tg β 1 ∓ tgα tgβ sin 2α = 2sin α cos α cotg 2α − 1 2tgα α tg2α = = cotg2 2cotgα 1 − tg 2α tg (α ± β ) =

sin α + sin β = 2sin

α +β

cos

α −β

2 2 α +β α −β cos α + cosβ = 2cos cos 2 2 1 sin α sin β = ( cos (α − β ) − cos (α + β ) ) 2 sin α cosβ =

1 ( sin (α + β ) + sin (α − β ) ) 2

900 + α cos α − sin α −cotgα − tgα

1800 − α sin α − cos α − tgα −cotgα

cos (α ± β ) = cos α cos β ∓ sin α sin β

cotgα cotgβ ∓ 1 cotgβ ± cotgα 2 2 cos 2α = cos α − sin α = 2 cos 2 α − 1 = 1 − 2sin 2 α 1 1 sin 2 α = (1 − cos 2α ) cos 2 α = (1 + cos 2α ) 2 2 cotg (α ± β ) =

sin α − sin β = 2sin

α −β 2

cos

α +β

2 α +β α −β cosα − cosβ = −2sin sin 2 2 1 cosα cosβ = ( cos (α − β ) + cos (α + β ) ) 2


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