TUGAS MATEMATIKA WAJIBNAMA : AJI TRI NUR LAELI1. a. 15 = 1 x 1 x 1 x 1 x 1 = 1b. (-2)3 = (-2) x (-2) x (-2) = (-8)

c. (−12 )4 = (−12 ) x (−12 )x (−12 ) x (−12 ) = 116

2. a. 22+5 = 27 = 2 x 2 x 2 x 2 x 2 x 2 x 2 b. 5 3+7 = 510 = 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5c. ( 23 )1+5 = ( 23 )6 = ( 23 ) x ( 23 )x ( 23 )x ( 23 ) x ( 23 )x ( 23 ) d. (−13 )2+3 = (−13 )5 = (−13 ) x (−13 )x (−13 ) x (−13 )x (−13 )e. 7 5+3 = 78 = 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7

3. a. 3 x2 yb. 5 x2

c. 2x3 y−23 z−2

= 2x3 1y213 z2

= ( 3 x3 z23 y2 )d. 4 x−2 y4

6 z−5 = ( 244 z53 x2 )

c. cda+b = cd ( a

−1+b1a+ 1b )

4. a. 5 x3+7 = 5 x10b. 10 x3 . 3x4 . x2 = 30x3+7+2 = 30x12c. 5ab x 2a2b = 10 a1+2 b1+1 = 10 a3 b2d. 2 x8 y4 . 5xy . 10x2y4 = 100x8+1+2 y4+1+4 = 100x11y9e. 3abc2 . 2a2 b3 c4 = 6a1+2 b1+3 c2+4 = 6a2 b4 c6

5. a. x9x3

= x9−3 = x6b. a5b3a2b2

= a5−2b3−2 = a3bc. a5b7 c2a3b4 c

= a5−3b7−4 c2−1 = a2b3cd. 4 x7 y82 x3 y2

= 2 x2−3 y3−2 = 2 x4 y6

e. 12x7 y53 y6

= 4 x7 y5−6 = 4 x7 y−1

PR1. sederhanakan bentuk akar di bawah ini:a. √ x6 c. √18a3b. √16 a3b d. √50a4b22. hitunglah nilainyaa. 813 c. 32−1

5 e. 8134b. 24315 d. 16−14

3. sederhanakanlah!a. 2532 c. 4√1296 e. √16a6b3 y4b. (−8 )−23

d √98 r3 y2 z4Jawaban

1. .a. √ x6 = x 62 = x3 c. √18a3 = 3√2 a 32b. √16a3b = 4 a b12 d. √50a4b2 = 5√2 a2b2. .a. 813 = 823 13 = 2 c. 32−1

5 = 25−15 = 2−1 c. 8134 = 34 34=33=27b. 24315=35 15=3 d. 16−14 =24− 14=2−1

3. .a. 2532 = 52 32=53=125b. (−8 )−2

3 = −23−23 = (−2)−2= 1

−2

c. 4√1296 = 6d. √98 r3 y2 z4 = √2.72 r3 y2 z4 = 7√2 r 32 y z2e. √98 r3 y2 z4= 4 a3 b32 y2

1. Rasionalkan 22√3 = 2

2√3 x 2√32√3 = 4 √3

4 √3 = 4 √312

= 1√33

= √33

2. Rasionalkan 4

5−√3 = 4

5−√3 x 5+√35+√3 = 20+4 √3

25−3 = 20+4 √3

22 = 10+2√3

11

3. Rasionalkan √3+√2√5−√3 = √3+√2

√5−√3 x √5+√3√5+√3 = √15+3+√10+√6

5−3 = √15+3+√10+√6

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