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Fisika
Ketinggian bola pada pemantulan ke-6:
h(6) = 35 – 6 × 26
= 3–1 × 26
= 1
3 × 64
= 64
3 cm
Jadi, ketinggian bola pada pemantulan ke-6
adalah 64
3 cm.
b. Tinggi bola pada pemantulan ke-10:
h(10) = 35 – 10 × 210
= 3–5 × 210
= 1
243 × 1.024
= 1.024
243 cm
Jadi, ketinggian bola pada pemantulan ke-10
adalah 1.024
243 cm.
Matematika Kelas X 5
8. Jawaban: c
45 − 28 − 3( 125 − 63)
= 9× 5 − 4× 7 − 3( 25× 5 − 9× 7)
= 3 5 – 2 7 – 3 (5 5 – 3 7)
= 3 5 – 2 7 – 15 5 + 9 7
= (3 −15) 5 + (−2 + 9) 7
= –12 5 + 7 7
9. Jawaban: c
(2 2 – 6 )( 2 + 6 )
= 2 2 ( 2 + 6 ) – 6 ( 2 + 6 )
= 2 2 2 + 2 2 6 – 6 2 – 6 6
= 2 × 2 + 2 2 2 3 – 3 2 2 – 6
= 4 + 4 3 – 2 3 – 6
= 2 3 – 2
= 2( 3 – 1)
10. Jawaban: a
x2 – y2 + 2xy
= (x + y)(x – y) + 2xy
= (2 – 3 + 2 + 3 )( 2 – 3 – 2 – 3 )
+ 2(2 – 3 )(2 + 3 )
= (4)(–2 3 ) + 2(2 – 3 )(2 + 3 )
= –8 3 + 2(4 – 3)
= –8 3 + 2
= 2 – 8 3
11. Jawaban: b
2 3 3 × 3 72 – 4 125 × 4 5
= 23 3 × 72 − 4 125 × 5
= 23 216 − 4 625
= 23 63 − 4 54
= 2 × 6 – 5
= 12 – 5
= 7
A. Pilihlah jawaban yang tepat.
1. Jawaban: b
Oleh karena 121 = 11 maka 121 bukan
merupakan bentuk akar.
2. Jawaban: b
48 + 243 = 16 × 3 + 81× 3
= 16 × 3 + 81 × 3
= 4 3 + 9 3
= 13 3
3. Jawaban: a
18 × 32 = 18 × 32 = 576 = 24
4. Jawaban: d
4 2 × 4 12 × 4 54 = 4 2 ×12 × 54
= 4 1.296
= 1.296
= 36 = 6
5. Jawaban: c
3 8.192 = 2 38.192 ×
= 6 4.096 × 2
= 6 46 × 2
= 46 2
6. Jawaban: e
3 50 − 8 + 128 − 5 18
= 3 25 × 2 − 4 × 2 + 64 × 2 − 5 9 × 2
= 3(5 2) − 2 2 + 8 2 − 5(3 2)
= (15 – 2 + 8 – 15) 2
= 6 2
7. Jawaban: a
1.100 − 5 44 + 275 − 2 11
= 100 ×11− 5 4 ×11+ 25 ×11− 2 11
= 10 11− 5 × 2 11+ 5 11− 2 11
= (10 – 10 + 5 – 2) × 11
= 3 11
6 Bentuk Pangkat, Akar, dan Logaritma
12. Jawaban: d
Diketahui 6 + 6 + 6 + . . . = x . . . (1)
Kedua ruas dikuadratkan diperoleh:
6 + 6 + 6 + 6 + . . . = x2 . . . (2)
Kurangkan (1) dari (2) diperoleh:
6 + 6 + 6 + 6 + . . . = x2
6 + 6 + 6 + . . . = x
––––––––––––––––––––––––– –
6 = x2 – x
⇔ x2 – x – 6 = 0
⇔ (x + 2)(x – 3) = 0
⇔ x = –2 atau x = 3
Oleh karena 6 + 6 + 6 + . . . mempunyai nilai
x dengan x berupa bilangan positif maka diambil
x = 3.
Jadi, nilai x adalah 3.
13. Jawaban: b
( 2 + 3 + 2 + 5)(− 2 + 3 + 2 − 5)
= (( 3 + 2) + ( 5 + 2 )) (( 3 + 2) − ( 5 + 2 ))
= ( )2 ( )2
3 + 2 − 5 + 2
= (3 + 4 3 + 4) – (5 + 2 10 + 2)
= 3 + 4 3 + 4 − 5 − 2 10 − 2
= −2 10 + 4 3
14. Jawaban: d
L = p ×
= (9 2 − 5 3) (3 2 + 3)
= 9 2 × 3 2 + 9 2 × 3 – 5 3 × 3 2
– 5 3 × 3
= 54 + 9 6 – 15 6 – 15
= 39 – 6 6
Jadi, luas persegi panjang (39 − 6 6) cm2.
15. Jawaban: e
Volume kubus
= r3
= (2 3 – 5)3
= (2 3 – 5)(2 3 – 5)(2 3 – 5)
= (12 – 4 15 + 5)(2 3 – 5)
= 24 3 – 8 45 + 10 3 – 12 5 + 4 75 – 5 5
= 24 3 – 8 × 3 5 + 10 3 – 12 5
+ 4 × 5 3 – 5 5
= 24 3 – 24 5 + 10 3 – 12 5 + 20 3 – 5 5
= 24 3 + 10 3 + 20 3 – 24 5 – 12 5 – 5 5
= 54 3 – 41 5
Jadi, volume kubus (54 3 – 41 5 ) cm3.
B. Kerjakan soal-soal berikut.
1. a. 2 175 – 5 343 – 63 – 3 112
= 2 25× 7 – 5 49× 7 – 9× 7 – 3 16 × 7
= 2 25 × 7 – 5 49 × 7 – 9 × 7
– 3 16 × 7
= 2 × 5 7 – 5 × 7 7 – 3 × 7 – 3 × 4 7
= 10 7 – 35 7 – 3 7 – 12 7
= –40 7
b. (5 27 – 6 2 )(6 27 – 8)
= 5 27 × 6 27 – 6 2 × 6 27
– 5 27 × 8 + 6 2 × 8
= 5 × 6 × 27 – 6 × 6 × 2 × 9× 3
– 5 × 9× 3 × 4× 2 + 6 × 2 × 4× 2
= 810 – 6 × 6 × 2 × 3 3 – 5 × 3 3 × 2 2
+ 6 × 2 × 2 × 2
= 810 – 108 6 – 30 6 + 24
= 834 – 138 6
c. 3 3 54 – 2 4 48 + 2 3 432 – 4 768
= 3 3 27 × 2 – 2 4 16 × 3 + 2 3 216 × 2
– 4 256 × 3
= 3 3 27 × 3 2 – 2 4 16 × 4 3
+ 2 3 216 × 3 2 – 4 256 × 4 3
= 3 × 3 × 3 2 – 2 × 2 × 4 3 + 2 × 6 × 3 2 – 4 × 4 3
= 9 3 2 – 4 4 3 + 12 3 2 – 4 4 3
= (9 + 12) 3 2 + (–4 – 4) 4 3
= 21 3 2 – 8 4 3
Matematika Kelas X 7
2. Misalkan AB dan AC merupakan sisi siku-siku dan
BC merupakan sisi miring.
AB2 = ( 5 + 3 − 2)2
= ( 5 + 3 – 2 )( 5 + 3 – 2 )
= 5( 5 + 3 − 2) + 3( 5 + 3 − 2)
− 2( 5 + 3 − 2)
= 5 + 15 − 10 + 15 + 3 − 6 − 10
− 6 + 2
= 5 + 3 + 2 + 15 + 15 – 10 – 10
– 6 – 6
= 10 + 2 15 – 2 10 – 2 6
AC2 = ( 3 − 5 + 2)2
= ( 3 − 5 + 2)( 3 − 5 + 2)
= 3( 3 − 5 + 2) − 5( 3 − 5 + 2)
+ 2( 3 − 5 + 2)
= 3 − 15 + 6 − 15 + 5 − 10 + 6
− 10 + 2
= 3 + 5 + 2 – 15 – 15 – 10 – 10 + 6 + 6
= 10 – 2 15 – 2 10 + 2 6
BC2= AB2 + AC2
= (10 + 2 15 – 2 10 – 2 6 ) + (10 – 2 15
– 2 10 + 2 6 )
= 10 + 10 + 2 15 – 2 15 – 2 10 – 2 10
– 2 6 + 2 6
= 20 – 4 10
= 4(5 – 10 )
BC2 = 4(5 − 10)
⇔ BC = 4(5 − 10)
= (2 5 − 10 ) cm
Jadi, panjang sisi miring segitiga tersebut
(2 5 − 10 ) cm.
3. a. Misalkan x = 4,25
10x = 42,5
100x = 425,5
100x – 10x = 425,5 – 42,5
⇔ 90x = 383
⇔ x = 383
90
Misalkan y = 1⎯,1
10y = 11⎯,1
10y – y = 11⎯,1 – 1⎯,1
⇔ 9y = 10
⇔ y = 10
9
42⎯,5 – 1⎯,1 = 383
90 – 10
9
= 383
90 – 100
90
= 283
90
b. Misalkan x = 3,412
⇔ 10x = 34,12
⇔ 1.000x = 3.412,12
1.000x – 10x = 3.412,12 – 34,12
⇔ 990x = 3.378
⇔ x = 3.378
990 = 563
165
Misalkan y = 1,06
⇔ 10y = 10,6
⇔ 100y = 106,6
100y – 10y = 106,6 – 10,6
⇔ 90y = 96
⇔ y = 96
90 = 16
15
3,412 : 1,0⎯6 = 563
165 : 16
15
= 563
165 × 15
16 = 563
176
4. a. (2 x + y )(3 x – 5 y ) x3y3
= (2 x × 3 x + y × 3 x – 2 x × 5 y
– y × 5 y ) x3y3
= (6x + 3 xy – 10 xy – 5y) x3y3
= (6x – 7 xy – 5y) xy xy
= 6x × xy xy – 7 xy × xy xy – 5y × xy xy
= 6x2y xy – 7x2y2 – 5xy2
h(6) = 35 – 6 × 26
= 3–1 × 26
= 1
3 × 64
= 64
3 cm
Jadi, ketinggian bola pada pemantulan ke-6
adalah 64
3 cm.
b. Tinggi bola pada pemantulan ke-10:
h(10) = 35 – 10 × 210
= 3–5 × 210
= 1
243 × 1.024
= 1.024
243 cm
Jadi, ketinggian bola pada pemantulan ke-10
adalah 1.024
243 cm.
Matematika Kelas X 5
8. Jawaban: c
45 − 28 − 3( 125 − 63)
= 9× 5 − 4× 7 − 3( 25× 5 − 9× 7)
= 3 5 – 2 7 – 3 (5 5 – 3 7)
= 3 5 – 2 7 – 15 5 + 9 7
= (3 −15) 5 + (−2 + 9) 7
= –12 5 + 7 7
9. Jawaban: c
(2 2 – 6 )( 2 + 6 )
= 2 2 ( 2 + 6 ) – 6 ( 2 + 6 )
= 2 2 2 + 2 2 6 – 6 2 – 6 6
= 2 × 2 + 2 2 2 3 – 3 2 2 – 6
= 4 + 4 3 – 2 3 – 6
= 2 3 – 2
= 2( 3 – 1)
10. Jawaban: a
x2 – y2 + 2xy
= (x + y)(x – y) + 2xy
= (2 – 3 + 2 + 3 )( 2 – 3 – 2 – 3 )
+ 2(2 – 3 )(2 + 3 )
= (4)(–2 3 ) + 2(2 – 3 )(2 + 3 )
= –8 3 + 2(4 – 3)
= –8 3 + 2
= 2 – 8 3
11. Jawaban: b
2 3 3 × 3 72 – 4 125 × 4 5
= 23 3 × 72 − 4 125 × 5
= 23 216 − 4 625
= 23 63 − 4 54
= 2 × 6 – 5
= 12 – 5
= 7
A. Pilihlah jawaban yang tepat.
1. Jawaban: b
Oleh karena 121 = 11 maka 121 bukan
merupakan bentuk akar.
2. Jawaban: b
48 + 243 = 16 × 3 + 81× 3
= 16 × 3 + 81 × 3
= 4 3 + 9 3
= 13 3
3. Jawaban: a
18 × 32 = 18 × 32 = 576 = 24
4. Jawaban: d
4 2 × 4 12 × 4 54 = 4 2 ×12 × 54
= 4 1.296
= 1.296
= 36 = 6
5. Jawaban: c
3 8.192 = 2 38.192 ×
= 6 4.096 × 2
= 6 46 × 2
= 46 2
6. Jawaban: e
3 50 − 8 + 128 − 5 18
= 3 25 × 2 − 4 × 2 + 64 × 2 − 5 9 × 2
= 3(5 2) − 2 2 + 8 2 − 5(3 2)
= (15 – 2 + 8 – 15) 2
= 6 2
7. Jawaban: a
1.100 − 5 44 + 275 − 2 11
= 100 ×11− 5 4 ×11+ 25 ×11− 2 11
= 10 11− 5 × 2 11+ 5 11− 2 11
= (10 – 10 + 5 – 2) × 11
= 3 11
6 Bentuk Pangkat, Akar, dan Logaritma
12. Jawaban: d
Diketahui 6 + 6 + 6 + . . . = x . . . (1)
Kedua ruas dikuadratkan diperoleh:
6 + 6 + 6 + 6 + . . . = x2 . . . (2)
Kurangkan (1) dari (2) diperoleh:
6 + 6 + 6 + 6 + . . . = x2
6 + 6 + 6 + . . . = x
––––––––––––––––––––––––– –
6 = x2 – x
⇔ x2 – x – 6 = 0
⇔ (x + 2)(x – 3) = 0
⇔ x = –2 atau x = 3
Oleh karena 6 + 6 + 6 + . . . mempunyai nilai
x dengan x berupa bilangan positif maka diambil
x = 3.
Jadi, nilai x adalah 3.
13. Jawaban: b
( 2 + 3 + 2 + 5)(− 2 + 3 + 2 − 5)
= (( 3 + 2) + ( 5 + 2 )) (( 3 + 2) − ( 5 + 2 ))
= ( )2 ( )2
3 + 2 − 5 + 2
= (3 + 4 3 + 4) – (5 + 2 10 + 2)
= 3 + 4 3 + 4 − 5 − 2 10 − 2
= −2 10 + 4 3
14. Jawaban: d
L = p ×
= (9 2 − 5 3) (3 2 + 3)
= 9 2 × 3 2 + 9 2 × 3 – 5 3 × 3 2
– 5 3 × 3
= 54 + 9 6 – 15 6 – 15
= 39 – 6 6
Jadi, luas persegi panjang (39 − 6 6) cm2.
15. Jawaban: e
Volume kubus
= r3
= (2 3 – 5)3
= (2 3 – 5)(2 3 – 5)(2 3 – 5)
= (12 – 4 15 + 5)(2 3 – 5)
= 24 3 – 8 45 + 10 3 – 12 5 + 4 75 – 5 5
= 24 3 – 8 × 3 5 + 10 3 – 12 5
+ 4 × 5 3 – 5 5
= 24 3 – 24 5 + 10 3 – 12 5 + 20 3 – 5 5
= 24 3 + 10 3 + 20 3 – 24 5 – 12 5 – 5 5
= 54 3 – 41 5
Jadi, volume kubus (54 3 – 41 5 ) cm3.
B. Kerjakan soal-soal berikut.
1. a. 2 175 – 5 343 – 63 – 3 112
= 2 25× 7 – 5 49× 7 – 9× 7 – 3 16 × 7
= 2 25 × 7 – 5 49 × 7 – 9 × 7
– 3 16 × 7
= 2 × 5 7 – 5 × 7 7 – 3 × 7 – 3 × 4 7
= 10 7 – 35 7 – 3 7 – 12 7
= –40 7
b. (5 27 – 6 2 )(6 27 – 8)
= 5 27 × 6 27 – 6 2 × 6 27
– 5 27 × 8 + 6 2 × 8
= 5 × 6 × 27 – 6 × 6 × 2 × 9× 3
– 5 × 9× 3 × 4× 2 + 6 × 2 × 4× 2
= 810 – 6 × 6 × 2 × 3 3 – 5 × 3 3 × 2 2
+ 6 × 2 × 2 × 2
= 810 – 108 6 – 30 6 + 24
= 834 – 138 6
c. 3 3 54 – 2 4 48 + 2 3 432 – 4 768
= 3 3 27 × 2 – 2 4 16 × 3 + 2 3 216 × 2
– 4 256 × 3
= 3 3 27 × 3 2 – 2 4 16 × 4 3
+ 2 3 216 × 3 2 – 4 256 × 4 3
= 3 × 3 × 3 2 – 2 × 2 × 4 3 + 2 × 6 × 3 2 – 4 × 4 3
= 9 3 2 – 4 4 3 + 12 3 2 – 4 4 3
= (9 + 12) 3 2 + (–4 – 4) 4 3
= 21 3 2 – 8 4 3
Matematika Kelas X 7
2. Misalkan AB dan AC merupakan sisi siku-siku dan
BC merupakan sisi miring.
AB2 = ( 5 + 3 − 2)2
= ( 5 + 3 – 2 )( 5 + 3 – 2 )
= 5( 5 + 3 − 2) + 3( 5 + 3 − 2)
− 2( 5 + 3 − 2)
= 5 + 15 − 10 + 15 + 3 − 6 − 10
− 6 + 2
= 5 + 3 + 2 + 15 + 15 – 10 – 10
– 6 – 6
= 10 + 2 15 – 2 10 – 2 6
AC2 = ( 3 − 5 + 2)2
= ( 3 − 5 + 2)( 3 − 5 + 2)
= 3( 3 − 5 + 2) − 5( 3 − 5 + 2)
+ 2( 3 − 5 + 2)
= 3 − 15 + 6 − 15 + 5 − 10 + 6
− 10 + 2
= 3 + 5 + 2 – 15 – 15 – 10 – 10 + 6 + 6
= 10 – 2 15 – 2 10 + 2 6
BC2= AB2 + AC2
= (10 + 2 15 – 2 10 – 2 6 ) + (10 – 2 15
– 2 10 + 2 6 )
= 10 + 10 + 2 15 – 2 15 – 2 10 – 2 10
– 2 6 + 2 6
= 20 – 4 10
= 4(5 – 10 )
BC2 = 4(5 − 10)
⇔ BC = 4(5 − 10)
= (2 5 − 10 ) cm
Jadi, panjang sisi miring segitiga tersebut
(2 5 − 10 ) cm.
3. a. Misalkan x = 4,25
10x = 42,5
100x = 425,5
100x – 10x = 425,5 – 42,5
⇔ 90x = 383
⇔ x = 383
90
Misalkan y = 1⎯,1
10y = 11⎯,1
10y – y = 11⎯,1 – 1⎯,1
⇔ 9y = 10
⇔ y = 10
9
42⎯,5 – 1⎯,1 = 383
90 – 10
9
= 383
90 – 100
90
= 283
90
b. Misalkan x = 3,412
⇔ 10x = 34,12
⇔ 1.000x = 3.412,12
1.000x – 10x = 3.412,12 – 34,12
⇔ 990x = 3.378
⇔ x = 3.378
990 = 563
165
Misalkan y = 1,06
⇔ 10y = 10,6
⇔ 100y = 106,6
100y – 10y = 106,6 – 10,6
⇔ 90y = 96
⇔ y = 96
90 = 16
15
3,412 : 1,0⎯6 = 563
165 : 16
15
= 563
165 × 15
16 = 563
176
4. a. (2 x + y )(3 x – 5 y ) x3y3
= (2 x × 3 x + y × 3 x – 2 x × 5 y
– y × 5 y ) x3y3
= (6x + 3 xy – 10 xy – 5y) x3y3
= (6x – 7 xy – 5y) xy xy
= 6x × xy xy – 7 xy × xy xy – 5y × xy xy
= 6x2y xy – 7x2y2 – 5xy2