PEMBAHASAN BENTUK PANGKAT DAN AKAR KELAS

PEMBAHASAN BENTUK PANGKAT DAN AKAR KELAS X

1. C
8



1
3

1

=

8

1
3

1 1
=3 =

√8 2

2. C
8
6

6 2
+
6 6

7 =7

6
6

2
6

1


1
3

3

3

=7 ×7 =7 ×7 =7×√ 7=7 √ 7

3. B

1
1
1
1
1
1
1
1
+

=
+
= q
+ p
= q p + q p
p−q
q− p
p
q
p
q
1+ x
1+ x
x
x
x x
x
x
x +x
x +x

1+ q 1+ p
+ q
+ p
q
p
q
x
x
x x
x
x
x
xp
xq
xp
xq+ x p
= q p + q p = q p =1
x +x
x +x
x +x

4. A

( x + y )3 a+1
=( x + y )3 a+1−( 2 a+5 )= ( x + y )3 a+1−2 a−5 =( x + y )a−4
2 a+5
(x + y )
5. B
3

a2 b
b4
a 2×3 b 3
b4 a 6 b3 ×b 4 1
×
=
×
= ×
× 6 3
c2
ac 3 c 2×3

ac 3 a 1
c ×c
5 7
1
a b
=a6−1 ×b 3+4 × 6+3 = 9
c
c

( )

(

)

6. B

( 3 x +3−x )2=( 6 )2
2
2

( 3 x ) +2 ( 3 x )( 3−x ) + ( 3−x ) =36
( 32 )x +2+ ( 32 )−x =36
9 x +2+9−x =36
9 x +9−x =36−2
9 x +9−x =34
1

7. D
2
(3 )2
3
31
1
=

=
9 ( 3x −2 )2 9
3 x −2




( )

1
3

()

2−2 x +4

⇔3
2
6−2 x=−
3
2
−2 x=− −6
3
−2−18
−2 x=
3

−20
−2 x=
3
−20
x=
−6
1
x=3
3

2

3

=3

9
1
⇔ 2 x−4 = 2
3

3

6−2 x

⇔3

1
3

( )

2


32
1
⇔ 2 x −4 = 2 ⇔ 32− ( 2 x−4 )=3 3
3
33


2

3

=3

8. D

1
1 9
3
3

√( ) (( ) )
3

8

1 9
3

( 3√ 2 )6

=

8

1 6
3

(2 )

9
9

8 8 8
= 6 = 2 = =2
2 4
3
2

9. B

√√ √(√ ) √( ) (( ) ) ( )
(( ) ) (√( )) ( )
3 2 27

8

=

3 2 27

1

27 2 27
=
=
8
8
8
3

27
=
8

1 1
3 2= 3

1 1
2 3=

27
8

1
2

27
8

1 1
×
2 3=

27
8

1 1
×
3 2

( )

3 12 √ 3
=
=
2
√2

√ 3 = √ 3 ×√ 2 = √ 6 = √ 6 = 1 √ 6
√ 2 √2 √ 2 √ 4 2 2
10. A

2

2

2
1 2

2

(a −a ) (a + a ) =[( a −a )(a +a )] =[(a ) −( a ) ]
1 2

2

1
2

1 2

2

1
2

1
2

1

2

1
2

1

2

1 2
2

1 2
1
1
1
−1 2
2
(
)
= a −a = a− =( a ) −2 ( a )
+

( a)

2

(a ) (a )

1 a4 −2 a2 +1
=a −2+ 2 = 2
a a
2
1
= 2 ( a2−1 )
a
2

11. B

p2 −2=( 1+ √ 3 ) −2=[ ( 12 ) +2 ( 1 ) ( √3 ) + ( √ 3 ) ] −2
=( 1+2 √3+3 )−2
=4 +2 √3−2=2+2 √ 3=2 ( 1+ √ 3 )=2 p
2

2

12. B

f ( x ) f ( y )=f ( x + y )
a x⋅a y =a x + y
a x + y =a x + y
13. A
3

(−z5 u5 ) =( −z 5 u5 )×( −z 5 u5 )×( −z 5 u5 )=−z 15 u15
14. C
6

x2
y6
:
y3
x−4

−3

x12 y−18
x 12
x12
x 12+12 x 24 24
:
=
×
=
= =x
y 18
x 12
y 18
y−18 y 18−18 y 0

( ) ( ) ( )( ) ( ) ( )
=

15. E



36
6
√0, 0036 = 10000 = 100 = 6 ×10 = 2 =0,2
0,3
3
3 100 3 10
10
10
16. D

3

82 x −1 =√ 4 2 x +3
3 2 x−1

(2 )

3 (2 x −1 )

2

6 x −3

=( 4

=4

1
2 x +3 2

)

1
(2 x+3 )
2

1
(2 x +3 )
22

2 =( 2 )
26 x −3 =( 2 )1 (2 x+3 )
26 x −3 =22 x +3
6 x−3=2 x+3
6 x−2 x=3+3
4 x =6
6 2 1
x= =1 =1
4 4 2
17. E

√ 3 √9 √ 27=3

x
y

√√

1
2



x
y

1
1
2
32

(

⇔ 3 9× ( 27 ) =3 ⇔ 3× 32× ( 3 )
1
1 1
2 2
32

1 1
3
2 2
2

( ( )) ( ( ))
( ( )) ( ( )) (

⇔ 3× 3 2×( 3 )
⇔ 3× 3

1 1
3
2+ 2 2
2

1
7
1+ 2
4

x
y

=3 ⇔ 3× 32 ×3

x
y

=3 ⇔ 3× 3
1
11
2
4

x
y

x
y

1 1
7
2 2
2

( ) =3 ⇔(3 ) =3 ⇔ 3

⇔3

11
8

) =3

=3

x
y

x
y

=3 ⇔ 3×3

=3

x
y

1
7
2
4

) =3

x
y

x
y

x=11
y=8
x+ y=19
18. A

5

2−n

n

−( 0,2 )
=
n
51−n + ( 0,2 )

5

2−n

5 1−n

n

( ) = 5 ⋅51 − 51 = 25⋅51 − 51 =24⋅51 =24 =4
1 1
1 1
1
6
1
5⋅ +
5⋅ +
6⋅
+( )
5
5 5
5 5
5

1

5

n

2

n

n

n

n

n

n

n

n

n

n

1

19. A
4

52 x+1−6⋅5 x +1=0
2x 1
x
5 ⋅5 −6⋅5 +1=0
2
5⋅( 5 x ) −6⋅5 x +1=0
x

Misalkan 5 =a
2
5 a −6 a+1=0
( 5 a−1 )( a−1 )=0
5 a−1=0 ∨ a−1=0
5 a=1
a=1
1
a=
5 x=1
5
1
5 x=
x=0
5
x=−1
20. A

93 x −2⋅3 3 x +1 −27=0
3x
( 32 ) −2⋅33 x⋅31 −27=0

( 33 x )2 −6⋅3 3 x−27=0
3x

Misalkan 3 =a
a2 −6 a−27=0
( a−9 ) ( a+3 )=0
a−9=0 ∨ a+3=0
a=9
a=−3
3x
2
3 =3
33 x =−3
3 x=2
x=tidak ada
2
x=
3
21. E
6 3
5

( 2 ) =2


18

5

=

1
2

18
5

=

1
3
3
5

2

=

1
23 ×2

3
5

1 1
= ×5
8 √8

22. A

3 √ 5×5 √5 3×5×√ 5× √5 3×5×5
=
=
=3
25
25
25
5

23. A



√10−√5 = √ 10 − √5 = 10 −1= 2−1=−1+ 2


√5
√ 5 √5 5
24. D

5
5
3 2+ 3 5 ( 3 √ 2+ √ 3 )
15 2+5 √ 3 15 √ 2+5 √ 3
=
× √ √ =
= √
=
2
2
15
3 √ 2−√ 3 3 √ 2− √ 3 3 √ 2+ √ 3 ( 3 √ 2 ) −( √ 3 ) 18−3
15 2 5 3
3
1
= √ + √ =√ 2+ √ = √ 2+ √ 3
15
15
3
3
25. A

√ 18+ √50−√ 72=√ 9⋅2+ √25⋅2−√ 36⋅2=3 √2+5 √ 2−6 √2=2 √ 2
26. A

√ 31+ √936−√ 21− √ 416= √31+√ 4×234−√ 21− √ 4×104
=√31+2 √ 234−√ 21−2 √104
=√ (18+13 ) +2 √( 18×13 )−√ ( 13+8 )−2 √( 13×8 )
=( √ 18+ √13 )−( √ 13− √8 )
=√18+ √ 8
=√ 9×2+ √ 4×2
=3 √2+2 √ 2=5 √2
27. B
2

2

√ 2−√ 3 = √2−√3 ×√ 2− √3 =( √ 2 ) −2 ( √2 ) ( √ 3 ) + ( √3 ) =2−2 √6+3 =5−2 √6
2−3
−1
√ 2+ √3 √2+ √3 √ 2− √3 ( √ 2 )2−( √ 3 )2
=−5+2 √6
a=−5
b=2
a+b=−5+2=−3
28. A
mn

1
m
p

√ √ a =( √ a )
p

n

1
1
m
p n

( ) =a

= (a )

p
mn

6

29. B

( m−7 √ 2 )( m+7 √2 ) =m2− ( 7 √2 )2 =m2 −98=( √ 18+ √ 80 )2 −98
2
2
=( √ 9⋅2+ √ 16⋅5 ) −98=( 3 √ 2+4 √5 ) −98
2
2
=( 3 √2 ) +2 ( 3 √ 2 )( 4 √5 )+ ( 4 √ 5 ) −98
=18+24 √ 10+80−98
=24 √ 10
30. C

2 √ 8+ √ 18+

1
1
√32+√ 200=2 √ 4⋅2+ √ 9⋅2+ √ 16⋅2+ √100⋅2
4
4
=4 √ 2+3 √ 2+ √ 2+10 √2=18 √2

31. C

√ √
3

3

3



3

3

49⋅ 49⋅√ 49⋅3 ⋯=a ⇔49⋅ 49⋅√ 49⋅3 ⋯=a3 ⇔ 49⋅a=a3 ⇔49=a 2 ⇔ 7=a




32. D

√ √

3 √ 24−2 √ 18 3 √ 24 2 √ 18
24
18
=

=−3
+2
=−3 √ 12+2 √ 9=−3 √ 4⋅3+2⋅3=−6 √ 3+6
2
2
−√ 2
−√ 2 − √ 2
33. D



( 5+ √ 2 )

Keliling =

2

2

x= ( 5+ √ 2 ) + ( 5−√ 2 ) =√( 25+10 √2+2 ) + ( 25−10 √ 2+2 )
=√ 25+25+10 √ 2−10 √ 2+2+2
=√ 54= √ 9×6=3 √ 6
( 5− √2 )
( 5+ √ 2 ) + ( 5−√2 ) +3 √6=10+3 √ 6

cm

34. E
L=4πr2

L=4⋅π⋅( 2 √ 2+ √ 6 )
2

2
2

=4⋅π⋅[ ( 2 √ 2 ) +2 ( 2 √2 )( √ 6 ) + ( √6 ) ]
=4⋅π⋅( 8+8 √3+6 )
2
=4⋅π⋅( 14+8 √3 ) =( 56+32 √ 3 ) π cm

7

35. B

f ( x+ 3 ) 2 x +3 2 x⋅23
21
= x −1 = x =2x⋅23 × x =23⋅21 =24 =f ( 4 )
f ( x−1 ) 2
2
2
21
36. D
2



3
31
=
x −2
9
3

( )
32

1
3

( )()
3

1
=
9

2 ( x −2 )

1

32
3
=( 3−2 )
2 x −4
3
3

2−2 x +4

2

3

=3
2
6−2 x=−
3
18−6 x=−2
−6 x=−20
20 10
x= =
6 3
37. B
5

3 x+3 =√ 27 x−5
3 x+3 =( 27
x+3

3 =3
x+3

1
x−5 5

)

3 ( x−5 )×

1
5

3
( x−5 )
5

3 =3
3
x+3= ( x−5 )
5
5 x+15=3 x−15
5 x−3 x=−15−15
2x=−30
x=−15

8

38. D

( 0, 25 )x +4 = √8 2 x−5
1
4

()

x +4

1
2 x−5 2

=( 8

)

1
3 ( 2 x−5 ) 2

−1 x +4

( 4 ) =( 2

)

1
6 x−15 2

2−2 ( x +4 )=( 2
−2 x−8

2

=2

)

15
3x−
2

15
2
−4 x−16=6 x−15
−4 x−6 x=−15+16
−10 x=1
1
x=− =−0,1
10
−2 x−8=3 x−

39. A

√ 108−

(

)

(

2
2
3+ 27
6+2 √ 27
6+2 √27
= √36×3−
× √ =6 √ 3−
=6 √ 3−
9−27
−18
3−√ 27
3−√ 27 3+ √27

(

)

)

6
2 √ 27
1 1
1 1
+
=6 √ 3+ + √ 9⋅3=6 √3+ + √ 9⋅3
−18 −18
3 9
3 9
1 1
19 3+1
=6 √ 3+ + √ 3= √
3 3
3
=6 √ 3−

(

)

40. C
−2

3

( 3 p−2 q 3 )
( 32 p−1 q 2 ) 32×3 p−1×3 q 2×3 3 6 p−3 q6 6−2 −3−(−4) 6−6 4
=
=
=
=3 p
q =3 p=81 p
( 32 p−1 q2 )−3 ( 3 p−2 q 3 ) 2 3 1×2 p−2×2 q 3×2 32 p−4 q 6

9