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SOLUSI SOAL OLIMPIADE MATEMATIKA Persiapan Olimpiade Sains Provinsi dan Nasional Tingkat SMA

Departemen Matematika - Wardaya College

MMXVIII-VII

1. 111 1

2x

2 111 1x 2 1 111x ( Kuadratkan kedua ruas)

2

2

2

2

(2 1) 111

4 4 1 111

4 4 110 0

2 2 55 0...(1)

x

x x

x x

x x

Kalikan (1) dengan 3x ……. 5 4 32 2 55 0x x x …(2)

Kalikan (1) dengan x ……… 3 22 2 55 0x x x …(3)

Kalikan (1) dengan 1 …….. 22 2 55 0x x …(4) Jumlahkan (2)(3)(4), maka diperoleh:

5 4 32 2 53 57 55 0x x x x 5 4 3 2004(2 2 53 57 54)x x x x = 5 4 3 2004(2 2 53 57 55 1)x x x x

2004

2004

(0 1)

( 1)

1

Jawaban : E

2. sin18 = 1 5

4

maka nilai 8 cba Jawaban : A

3. 39

9)(

x

x

xf

xx

x

xf93

3

39

9)1(

1

1

11)( xfxf

1996

998

1996

999

1996

997...

1996

1995

1996

1

1996

1995...

1996

2

1996

1ffffffff

2

1995

33

3997

2

1997.1

1996

1995...

1996

2

1996

1

ffff

Jawaban : A



4. Misalkan : 2x a , 2y b , 2z c . Substitusikan ke soal semula, sehingga

diperoleh : 2

2

2

42.........(1)

6...........(2)

30.......(3)

a bc

b ac

c ab

(1) – (2) : 2

2

2 2

42.........(1)

6...........(2)

36

( )( ) ( ) 36

( )( ) 36

36( ) .....(4)

( )

a bc

b ac

a b bc ac

a b a b c a b

a b a b c

a b ca b

(1) - (3) : 2

2

2 2

42.........(1)

30.......(3)

72

( )( ) ( ) 72

( )( ) 72

72( ) .....(5)

( )

a bc

c ab

a c bc ab

a c a c b a c

a c a b c

a b ca c

Substitusikan (4) dengan (5) :

36 72

( ) ( )a b a c

36( ) 72( )

( ) 2( )

2 2

2

....(6)2

a c a b

a c a b

a c a b

b a c

a cb

Substitusikan (6) ke (2) : 2

2

2 2

2 2

2 2

2

6...........(2)

( ) 62

26

4

2 4 24

2 24

( ) 24

( ) 24.....(7)

b ac

a cac

a ac cac

a ac c ac

a ac c

a c

a c

Substitusikan (6) ke (1) : (7) + (8) :

( ) 2 6.....(7)

(2 ) 7 6.....(8)

3 9 6

3 6

a c

a c

a

a



2

2

22

2 2

42.........(1)

( ) 422

422

2 84

(2 )( ) 84

(2 ). 24 84

(2 ) 7 6........(8)

a bc

a ca c

ac ca

a ac c

a c a c

a c

a c

( ) 24.....(7)

3 6 2 6

6

a c

c

c

2

2

42.........(1)

(3 6) 6 42

54 6 42

6 12

2 6

a bc

b

b

b

b

Sesuai dengan pemisalan awal : 2

2(3 6)

54

x a

2

2(2 6)

24

y b

2

2( 6)

6

z c

54

24

6

x

y

z

maka 84 zyx

Jawaban : C

5. (√5 + 2 + √3 + √6)( √5 - √3 + 2 - √6)

= 5 -√15 + 2√5 - √30 + 2√5 - 2√3 + 4 - 2√6 + √15 - 3 + 2√3 - √18 + √30 - √18 + 2√6 – 6

= 5 + 4√5 + 4 – 3 - 2√18 - 6

= 4√5 - 6√2

(4√5 - 6√2)( 7√2 + 2√5)

= 28√10 + 40 – 84 - 12√10

= 16√10 - 44 Jawaban : A

6. baxba

xa

ab

xa

ab

ab

0,tan.

sin.1

sin.

.1 2



.cos

sin

cos

sin).(.

sin).(

sin

cos

sin.)sin1(.

sin).(

sin

cos

sin.cos..

sin).(

sin.

.1

2

2

2

22

2

2

22

2

x

x

x

xaba

xaba

x

x

xbxa

xaba

x

x

xbxa

a

xaba

xa

ab

ab

Jawaban : E

7. Ingkaran dari semua adalah ada Jadi ingkaran dari kalimat “ Semua anak – anak suka bermain air” adalah “Ada anak – anak yang tidak suka bermain air” Jawaban : C

8. QT = (𝑥 −3

2𝑦 4)

PQT = R

(12 40 −11

) (𝑥 −3

2𝑦 4) = (

96 −2066 −44

)

(𝑥 −3

2𝑦 4) = -

1

132(

−11 −40 12

) (96 −2066 −44

)

= (

11

132

4

132

0 −12

132

) (96 −2066 −44

)

= (

1056

132+

264

132−

220

132−

176

132

0 − 792

1320 +

520

132

)

= (10 −3−6 4

)

x = 10 2y = -6 y = -3 2x + y = 2(10) – 3 = 17 Jawaban : E

9. 100+10+1+0,1+0,01+0,001+...=9

1111

Jawaban : E 10. 4x – 12.2x + 32 = 0

22x – 12.2x + 32 = 0 Misal : 2x = a a2 – 12a + 32 = 0 ( a – 8 ) (a – 4 ) = 0 a = 8 a = 4 2x = 23 2x = 22 x = 3 x = 2



x1x2 = 3 . 2 = 6 Jawaban : B

11. 4 1 -11 30 -8 4 -28 8

1 7 2 0 ( x – 4 ) (x2 – 7x + 2)

Jawaban : D

12.

Jika

2

0

20072007

2007

cossin

sin

dxxx

x kita misalkan dengan I, maka

4

2

2cossin

cossin

2

0

2

0

20072007

20072007

I

I

Idxxx

x

x

x

Jadi, 14

4

cossin

sin4

2

0

20072007

2007

dxxx

x

.

Jawaban : D

13. Misal : 1

𝑥 = a ,

1

𝑦 = b ,

1

𝑧 = c

2a + 4b + 7c = 31 3a + 2b + 5c = 22 a + 3b + 4c = 19

2

020072007

20072

0

20072007

2007

)2

(cos)2

(sin

)2

(sin

cossin

sin

dx

xx

xdx

xx

x

2

0

20072007

2007

sincos

cos

dxxx

x



2a + 4b + 7c = 31 . 3 6a + 12b + 21c = 93 3a + 2b + 5c = 22 . 2 6a + 4b + 10c = 44 - 8b + 11 c = 49 2a + 4b + 7c = 31 .1 2a + 4b + 7c = 31 a + 3b + 4c = 19 .2 2a + 6b + 8c = 38 –

-2b – c = -7

8b + 11c = 49 .1 8b + 11 c = 49

-2b – c = -7 .4 -8b – 4c = -28 +

7c = 21 → c = 3

-2b – 3 = -7

-2b = -4→ b = 2

a + 3(2) + 4(3) = 19

a + 6 + 12 = 19 → a = 1

1

𝑥 = a

1

𝑦 = b

1

𝑧 = c

1

𝑥 = 1

1

𝑦 = 2

1

𝑧 = 3

𝑥 = 1 y = 1

2 z =

1

3

x + y + z = 1 + 1

2 +

1

3 =

6+3+2

6 =

11

6

Jawaban : B

14. x + 1

𝑥 = √3

(x + 1

𝑥)3 = x3 + 3x2.

1

𝑥 +3x.(

1

𝑥)2 + (

1

𝑥)3

= x3 + 3x + 3

𝑥 + (

1

𝑥)3

x3 + (1

𝑥)3 = (x +

1

𝑥)3 – 3x -

3

𝑥

= (x + 1

𝑥)3 – 3(x +

1

𝑥)

= (√3)3 – 3(√3)

= 3√3 - 3√3 = 0 Jawaban : A

15. prinsip: ))(()( 22 bababa

000.000.000.000.000.554.555.555.555.555

)554.555.555.555.555)(000.000.000.000.000.1(

223.222.222.222.222777.777.777.777.777 22

Maka, jumlah semua angkanya adalah (5x14)+(4x1) = 74 Jawaban : C

G H



16.

Lihat segitiga OCG

Maka OC = 1

2𝑎√2 , CG = a

tan α = 𝑂𝐶

𝐶𝐺 =

1

2𝑎√2

𝑎 =

1

2√2

Jawaban : A

17. a + b = -c b + c = -a c + a = -b

maka (𝑎+𝑏)(𝑏+𝑐)(𝑐+𝑎)

𝑎𝑏𝑐 =

−𝑐.−𝑎.−𝑏

𝑎𝑏𝑐 =

− 𝑎𝑏𝑐

𝑎𝑏𝑐 = -1

Jawaban : B

18. 𝑎

𝑏 +

𝑎+10𝑏

𝑏+10 𝑎 = 2

x b (b+10a)

a(b+10a) + b(a+10b) = 2b(b+10a)

ab +10a2 + ab + 10b2 = 2b2 + 20ab

10a2 -18ab + 8b2 = 0 : b2

10(𝑎

𝑏)2 – 18

𝑎

𝑏 + 8 = 0

: 2

5(𝑎

𝑏)2 – 9

𝑎

𝑏 + 4 = 0

(5(𝑎

𝑏) - 4)( (

𝑎

𝑏) - 1)= 0

𝑎

𝑏 =

4

5 atau

𝑎

𝑏 = 1

Karena dikatakan a dan b adalah bilangan berbeda maka 𝑎

𝑏 =

4

5

Jawaban : B

19. ( 1 - 1

4 ) ( 1 -

1

5 ) ( 1 -

1

6 ) ... ( 1 -

1

100 )

= 3

4 .

4

5 .

5

6 .....

99

100

= 3

100

Jawaban : C

20. 𝑎𝑏

𝑐 .

𝑐𝑎

𝑏 = a2 = 1 . 3 = 3

A B

C D

E F

O

α

O C

G

α



𝑎𝑏

𝑐 .

𝑏𝑐

𝑎 = b2 = 1 . 2 = 2

𝑏𝑐

𝑎 .

𝑐𝑎

𝑏 = c2 = 2 . 3 = 6

Maka a2 + b2 + c2 = 3 + 2 + 6 = 11 Jawaban : C

21. 1! = 1 2! = 2 . 1 = 2 3! = 3 . 2 . 1 = 6 4! = 4 . 3 . 2 . 1 = 24 5! = 5 . 4 . 3 . 2 . 1 = 120 Jumlah satuan = 1 + 2 + 6 + 4 + 0 = 13 Jadi angka satuannya = 3 Jawaban : D

22. 𝐶3

4 − 𝐶2 6

𝐶5 10 =

4 .15

252 =

60

252 =

5

21

Jawaban : A

23.

a √𝑎2 + 1

α

1

(sinα + cosα)2 = sin2α + cos2α + 2sinαcosα

= 1 + 2 𝑎

√𝑎2+ 1

1

√𝑎2+ 1

= 1 + 2𝑎

𝑎2 + 1

= 𝑎2+ 2𝑎+1

𝑎2+1

Jawaban : B

24.

A a C O 11

11

7

7

D E



DE = DC + CE = 22 + a

OD = OE = 1

2 (22 + 𝑎) = 11 +

1

2𝑎

AO = OD – DA = (11 + 1

2𝑎) - 11 =

1

2𝑎

Sehingga OC = 11 – 1

2𝑎

Perhatikan ∆ ACB DAN OCB yang mempunyai sisi CB yang sama

(18)2 – (11)2 = (4 + 1

2𝑎)2 – (11 -

1

2𝑎)2

324 – 121 = 16 + 4a + 1

4𝑎2 – 121 + 11a -

1

4𝑎2

203 = 15a – 105 308 = 15a

a = 208

15

Jawaban : A



25.

0,21

0,14 0,14

0,21

X . y = 2

Bilangan 1 : x → x – o,28

Bilangan 2 : 2

𝑥 → (

2

𝑥 - 0,42)

Luas = (x – o,28) (2

𝑥 - 0,42)

Luas = 2 – 0,42x – 0,56

𝑥 + 0,1176

Luas = – 0,42x – 0,56

𝑥 + 2,1176

Luas’ = o = -0,42 + 0,56

𝑥2

0,56

𝑥2 = 0 , 42

𝑥2 = 0,56

0,42 =

9

3

x = 2

3 √3 , y = √3

Jawaban : A

26. ∫𝑥

𝑥+1 dx = ∫

𝑥+1−1

𝑥+1 dx

= ∫𝑥+1

𝑥+1 dx - ∫

1

𝑥+1 dx

= ∫ 𝑑𝑥 - ∫ 1

𝑥+1 dx

= x – ln |x + 1| + c

Jawaban : C

27. | 2x-5 | < | x+4 | ( 2x-5 )2 < ( x + 4 )2 4𝑥2 - 20x + 25 < 𝑥2 + 8x + 16 3𝑥2 – 28x + 9 < 0 ( 3x -1 ) ( x – 9 ) < 0 3x – 1 = 0 x – 9 = 0

x = 1

3 x = 9

Hp = {x | 1

3 < x < 9 }

Jawaban : E

9 1

3



28. √2,5+ √0,4

√10 − √2,5+ √0,4 .

√10

√10 =

√25+ √4

√100 − √25+ √4

= 5+2

10−5+2 =

7

7 = 1

Jawaban : A

29. x + √𝑥 = 1 → √𝑥 = 1 – x

(√𝑥 = 1 – x )2 x = 1 – 2x + 𝑥2 : x

1 = 1

𝑥 - 2 + x

x + 1

𝑥 = 3

Jawaban : D

30. Kenaikan bakso = 16

100 . 5000 = 800

Harga bakso menjadi = Rp. 800,- + Rp. 5.000,- = Rp. 5.800,-

Kenaikan jus = 4

100. 5000 = 200

Harga jus menjadi = Rp. 200,- + Rp. 5.000,- = Rp. 5.200,- Harga bakso + jus = Rp. 11.000,- Selisih kenaikan = Rp. 11.000,- - Rp. 10.000,- = Rp. 1.000,-

Persen kenaikan = 1000

10000 . 100% = 10%

Jawaban : B

31. (𝑠𝑖𝑛4 750 - 𝑐𝑜𝑠4 750) + (𝑠𝑖𝑛4 750 + 𝑐𝑜𝑠4 750) (𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 - 2𝑠𝑖𝑛2 75 𝑐𝑜𝑠2 750)

(𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 - 1

2 .4𝑠𝑖𝑛2 75 𝑐𝑜𝑠2 750)

(𝑠𝑖𝑛2 750 - 𝑐𝑜𝑠2 750) (𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750) ((𝑠𝑖𝑛2 750 + 𝑐𝑜𝑠2 750)2 -1

2 (2𝑠𝑖𝑛750 𝑐𝑜𝑠 750)2)

− 𝑐𝑜𝑠 1500 . 1 ( 1 - 1

2(sin 1500)2)

1

2√3 ( 1-

1

8 ) =

1

2√3 .

1

8 =

7

16 √3

Jawaban : A

32. Perhatikan ∆BCD

Sin 300 = 1

𝐵𝐷

1

2 =

1

𝐵𝐷

BD = 2

BC = √22 − 12 = √3

BR = 1

2BA =

1

2√3

B

A C

D

T 30o Ѳ



CR = √(√3)2 − (1

2√3)

2

= 3

2

Jadi , tan Ѳ = 𝐶𝐷

𝐶𝑇 =

13

2

= 2

3

Jawaban : D

33.

Misalkan AB = 2 , maka BP = √5 dan BG = 2√2 Aturan kosinus (PG)2 = (PB)2 + (BG)2 – 2.PB.BG. cosѲ

1 = 5 + 8 - 2√5 . 2√2 cos Ѳ

1 = 13 - 4√10 cos Ѳ

Cos Ѳ = 3

√10

√10 1

3

Jadi , tan Ѳ = 1

3

Jawaban : C 34. Persamaan garis melalui titik (7,1) adalah

y – 1 = m ( x – 7 ) y = mx + 1 – 7m , maka c2 = R2 ( 1 + m2 ) ( 1 – 7m )2 = 25 ( 1 + m2 ) 1 – 14m + 49m2 = 25 + 25m2 24m2 – 14m – 24 = 0 ( 4m + 3 ) ( 3m – 4m )= 0

m1 = - 3

4 , m2 =

4

3

Untuk m1 = - 3

4 Untuk m2 =

4

3

y = - 3

4x + 1 – 7 (-

3

4) y =

4

3x + 1 – 7 (

4

3 )

4y = -3x + 4 +21 3y = 4x + 3 - 28 3x + 4y = 25 4x – 3y = 25 Jawaban : B

35. Karena akar – akar membentuk deret aritmatika dengan beda 2 , maka : Misal : x1 = p , x2 = p + 2 , x3 = p + 4 , x4 = p + 6

x1 + x2 + x3 + x4 = - 𝑏

𝑎

p + ( p + 2 ) + ( p + 3 ) + ( p + 4 ) = 8 4p + 12 = 8

A B

C D

E

G H

F

Ѳ

P Q

Ѳ



Maka p = - 1 Dengan demikian , maka x1 = -1 , x2 = 1 , x3 = 3 , x4 = 5

x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4 = 𝑐

𝑎

(-1)(1) + (-1)(3) + (-1)(5) + (1)(3) + (1)(5) + (3)(5) = 𝑎

1 , → a = 14

x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 = - 𝑑

𝑎

(-1)(1)(3) + (-1)(1)(5) + (-1)(3)(5) + (1)(3)(5) = − (−𝑏)

1 , → b = - 8

x1x2x3x4 = 𝑐

𝑎

(-1)(1)(3)(5) = 𝑐

1 , → c = - 15

Jawaban : E

36. Misalkan x + y = a , x – y = b , maka :

a + √𝑎 = 30

a + √𝑎 - 30 = 0

(√𝑎 + b ) (√𝑎 - 5 ) = 0

√𝑎 = - 6 (TM) , √𝑎 = 5

b + √𝑏 = 12

b + √𝑏 - 12 = 0

(√𝑏 + 4 )( √𝑏 - 3 ) = 0

√𝑏 = - 4 (TM) , √𝑏 = 3

Maka nilai dari √𝑥2 − 𝑦2 = √(𝑥 + 𝑦)(𝑥 − 𝑦)

= √𝑥 + 𝑦 . √𝑥 − 𝑦

= √𝑎 . √𝑏

= 5 . 3 = 15

Jawaban : D

37. Misalkan : x! =a maka : 𝑎!

𝑎 = 120

𝑎 .(𝑎−1)!

𝑎 = 120

a ( a – 1 )! = 120a ( a – 1 )! = 5! a – 1 = 5 a = 6 x! = a x! = 6 x = 3

sin𝜋𝑥

2 = sin

3𝜋

2 = -1

Jawaban : A

38. 𝑏

𝑎 = 2005

𝑐

𝑏 = 2005

b = 2005a c = 2005b b = 2005a



c = 2005b + b + c = 2005a + 2005b b + c = 2005 ( a + b ) 𝑏+𝑐

𝑎+𝑏 = 2005

Jawaban : A 39. Misalkan : a = 𝑥2 - 10x – 29

Maka : 𝑥2 - 10x – 45 = a – 16 𝑥2 - 10x – 69 = a – 40 1

𝑎 +

1

𝑎−16 -

2

𝑎−40 = 0

(a – 16) (a – 40) + a (a – 40) – 2a(a – 16) = 0 (a2 – 56a + 640) + (a2 – 40a) – 2a2 + 32a = 0 -56a + 640 – 40a + 32a = 0 64a = 640 a = 10 10 = x2 – 10x – 29 x2 – 10x – 39 = 0

x + y = - 𝑏

𝑎 = 10

Jawaban : B 40. 432 = 264

(44)10 = 280 1618 = 272 (83)8 = 272 Yang paling besar adalah 281 Jawaban : A

41. Bila keterangan – keterangan dinyatakan dalam diagram venn , maka diagramnya adalah Tring

Trang

Trung

Jadi , yang paling benar adalah Y dan Z

Jawaban : E

42. 5 ekor ~ 5 lapangan bola dalam 5 hari 1 ekor ~ 1 lapangan bola dalam 5 hari Jadi , 3 ekor kambing dapat menghabiskan rumput seluas 3 kali lapangan bola dalam 5 hari. Jawaban : D

43. (2x + 𝑦

2 )-1 . [ (2x)-1 + (

𝑦

2)-1 ] =

1

2𝑥+ 𝑦

2

. (1

2𝑥 +

1𝑦

2

)

= 1

2𝑥+ 𝑦

2

(1

2𝑥 +

2

𝑦)



= 1

2𝑥+ 𝑦

2

(𝑦+4𝑥

2𝑥𝑦)

= 1

2𝑥+ 𝑦

2

(𝑦

2+ 2𝑥

𝑥𝑦)

= 1

𝑥𝑦 = (xy)-1

Jawaban : D

44. 1

𝑚 +

1

𝑛 =

4

7 →

1

𝑚 +

1

𝑛 =

8

14

1

𝑚 +

1

𝑛 =

7

14 +

1

14

1

𝑚 +

1

𝑛 =

1

2 +

1

14

Didapat m = 2 , n = 14 Jadi m2 + n2 = 22 + 142 = 4 + 196 = 200

Jawaban : D 45. C

750 E

A 550 D B

Sudut B = 1800 – (550 + 750) = 500

Karena BD = BE , maka sudut D = sudut E

Sudut D + sudut E + 500 = 1800

Sudut E = 650

Jadi Sudut BED = 650

Jawaban : C

46. C 3x

X 2x

A B

x0 + 2x0 + 3x0 = 1800

x0 = 300

𝐴𝐵

sin 3𝑥 =

𝐵𝐶

sin 𝑥

𝐴𝐵

𝐵𝐶 =

sin 3𝑥

sin 𝑥 =

sin 90

sin 30 =

112

= 2

Maka AB : BC = 2 : 1

Jawaban : C

47. y = 𝑥−1

2𝑥+3



2xy + 3y = x – 1 2xy – x = - 3y -1 (2y – 1)x = -3y -1

x = −3𝑦−1

2𝑦−1 =

1+3𝑦

1−2𝑦

Jawaban : A

48. Misalkan keliling = 12x 4x 4x 4x

L = 1

2 . 4x . 4x sin 600

= 1

2 . 4x . 4x .

1

2√3

= 4x2√3 3x 3x 3x 3x

L = 3x.3x = 9x2

Keliling = 2𝜋R = 12x

= R = 6𝑥

𝜋

Luas = 𝜋R2

= 𝜋.36𝑥2

𝜋2

= 11,1x2

Jawaban : C

49. x2 ≥ 4 |x – 1| ≤ 2 x2 – 4 ≥ 0 ( x – 1 )2 ≤ 4 ( x+2 )(x – 2) ≥ 0 x2 – 2x – 3 ≤ 0 x1 = 2 , x2 = -2 ( x – 3 ) ( x + 1 ) ≤ 0 x1 = 3 , x2 = -1

+ + + + - -



Yang memenuhi kedua – duanya adalah 2 ≤ x ≤ 3 Jawaban : B

50.

OD = jari – jari lingkaran dalam ∆ABC

= √𝑠(𝑠−𝑎)(𝑠 − 𝑏)(𝑠−𝑐 )

𝑠

= √27(27−12)(27 − 24)(27−18 )

27

= 27

27√15 = √15

AB2 = BC2 + AC2 – 2BC.AC cos C 144 = 576 + 324 – 2.24.18 cos C 2.24.18 cos C = 756

cos C = 7

8

sin C = √1 − 𝑐𝑜𝑠2𝐶

= √1 − 49

64

= √15

8

Perhatikan ∆NPC

sin C = 𝑁𝑃

𝑁𝐶 =

√15

8

√15

𝑁𝐶 =

√15

8

Maka NC = 8 AN = 18 – 8 = 10

Keliling ∆AMN = 𝐴𝑁

𝐴𝐶. Keliling ∆ABC

= 10

18 (54)

= 30 Jawaban : C

-2 2 1 3

C P

O

D B

A

N M

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