suku aljabar\lampiran-lampiran\Lampiran 6 INSTRUMEN SOAL TES(aljabar)
Lampiran 6
INSTRUMEN TES
NILAI
LEMBAR SOAL TES
Nama Sekolah :
Nama Siswa
Hari/ Tanggal :
Kelas/ No. Absen :
Mata Pelajaran : Matematika
Alokasi Waktu
Kerjakan soal di bawah ini dengan benar dan jelas!
1. Selesaikan operasi penjumlahan berikut!
a.
( x 2+ 3 y ) + (−7 y + 4 x 2−5 )
( x+ 2 y −z )+ (2 x− y +3 z )
b.
2. Selesaikan operasi pengurangan berikut!
7 p−8−q−5 p−2 q−3
a.
3
3
b.
3 m n−8 m−7 n−12−m n−27 m−5 n
3. Selesaikan operasi kali berikut!
( 3 x−5 ) ( 2 x +6 )
a.
b.
( 4 x2 − y ) ( x +2 x−7 )
4. Selesaikan operasi bagi berikut!
3 2
a.
64 a b ∶ 4 ab
b.
28 a4 b c3 ∶ ( 2 abc × 7 a2 c )
5. Selesaikan operasi pangkat berikut!
a.
( 4 a+7 b )2
b.
( 9 p−5 p )3
6. Selesaikan operasi berikut!
a.
( 4 x +1 )2− (2 x−2 )2
3
b.
( 4 a+b ) −( 48 a4 b3 c ∶ ( 8 a b2 ×2 bc ) )
:
: 2 x 45 menit
KUNCI JAWABAN TES
1. a. ( x 2+ 3 y ) + (−7 y + 4 x 2−5 )
skor
= ( x 2+ 4 x 2 ) + ( 3 y−7 y ) −5 ....................................
1
= ( 1+4 ) x 2 + ( 3−7 ) y−5 ........................................
1
= 5 x2 −4 y −5 .............................................................
1
b. ( x+ 2 y −z )+ (2 x− y +3 z )
= ( x+ 2 x ) + ( 2 y− y ) + (−z+3 z ) .............................
1
= ( 1+2 ) x + ( 2−1 ) y + (−1+3 ) z .............................
1
= 3 x+ y +2 z .................................................................
1
Skor maksimum:
6
2. a. 7 p−8−q−5 p−2 q−3
= ( 7 p−5 p ) + (−q−2 q ) + (−8−3 ) ...............................
1
= ( 7−5 ) p+ (−1−2 ) q+ (−11 ) ......................................
1
=
2 p−3 q−11 ....................................................................
1
b. 3 m3 n−8 m−7 n−12−m3 n−27 m−5 n
= ( 3 m3 n−m3 n ) + (−8 m−27 m )+ (−7 n−5 n )−12 ...............................................
1
= ( 3−1 ) m3 n+ (−8−27 ) m+ (−7−5 ) n−12 ..........................................................
1
= 2 m3 n−35 m−12n−12 ............................................................................................ 1
Skor maksimum:
3. a. ( 3 x−5 ) ( 2 x +6 )
6
= 3 x ( 2 x+6 )−5 ( 2 x+6 ) ................................................................
1
= 3 x ( 2 x )+3 x ( 6 )−5 ( 2 x )−5 ( 6 ) ...................................................
1
= 6 x 2+18 x−10 x−30 .................................................................. 1
= 6 x 2+ (18 x−10 x )−30 ...............................................................
1
= 6 x 2+ (18−10 ) x−30 .................................................................
1
= 6 x 2+ 8 x−30 ................................................................................
1
b. ( 4 x2 − y ) ( x +2 x−7 ) = 4 x 2 ( x+2 x−7 )− y ( x+2 x−7 ) .........................................
1
= 4 x 2 ( x )+ 4 x 2 ( 2 x )+ 4 x 2 (−7 )− y ( x ) − y ( 2 x )− y (−7 ) ....
1
= 4 x 3 +8 x 3−28 x 2−xy−2 xy +7 y .....................................
1
= ( 4 x3 +8 x 3 )−28 x 2 + (−xy−2 xy )+ 7 y ...........................
1
= ( 4 +8 ) x 3−28 x 2+ (−1−2 ) xy +7 y ..................................
1
= 12 x 3−28 x 2−3 xy +7 y ....................................................... 1
4.
Skor maksimum
12
3 2
a. 64 a b ∶ 4 ab
=
3
2
64 a b
4 ab
..................................................................................................
1
2
4 ab × 16 a b
4 ab
=
...........................................................................................
=
1
2
16 a b .....................................................................................................
1
b. 28 a4 b c3 ∶ ( 2 abc × 7 a2 c )
=
28 a4 b c 3
2 abc ×7 a 2 c
........................................................................
=
............................................................................
1
28 a4 b c 3
14 a3 b c 2
1
3
2
14 a b c × 2ac
3
2
14 a b c
=
...................................................................
1
2 ac
=
....................................................................................
1
Skor maksimum:
7
5. a. ( 4 a+7 b )2
2
=
1
1
2
..............................................................
1
= 1 ( 16 a2 ) +2 ( 4 a ) (7 b ) +1 ( 49 b2 ) .............................................................
1
1 ( 4 a ) +2 ( 4 a ) ( 7 b ) +1 (7 b )
= 16 a2 +8 a ( 7 b )+ 49 b2 .............................................................................. 1
= 16 a2 +56 ab+ 49 b2 ................................................................................. 1
b. ( 9 p−5 p )3
3
=
2
1
1
2
1 ( 9 p ) +3 ( 9 p ) (−5 p ) +3 ( 9 p ) (−5 p ) + 1 (−5 p )
3
...........................
1
= 1 ( 729 p3 ) +3 ( 81 p2 ) (−5 p ) +3 ( 9 p ) ( 25 p 2 ) +1 ( −125 p3 ) ....................
1
= 729 p 3+243 p2 (−5 p ) +27 p ( 25 p2 )−125 p3 .........................................
1
= 729 p 3−1215 p 3+ 675 p3 −125 p3 .........................................................
1
= 64 p3 ............................................................................................................
1
Skor maksimum
9
6. a. ( 4 x +1 )2− (2 x−2 )2
=
( 1 ( 4 x )2 +2 ( 4 x )1 (1 )1 +1 ( 1 )2 )−( 1 ( 2 x )2+2 ( 2 x )1 (−2 )1 +1 (−2 )2)
=
( 1 ( 16 x 2) + 2 ( 4 x )( 1 ) +1 ( 1 ) )−( 1 ( 4 x 2 ) +2 ( 2 x ) (−2 ) +1 ( 4 ) )
=
( 16 x 2+ 8 x ( 1 ) +1 )−( 4 x 2 +4 x (−2 )+ 4 ) ...............................................................
..................
.............................
1
1
1
= 16 x 2+8 x +1−4 x 2+8 x−4 ...................................................................................
1
= ( 16 x 2−4 x 2 ) + ( 8 x +8 x ) + ( 1−4 ) ..........................................................................
1
= ( 16−4 ) x 2 + ( 8+8 ) x + (−3 ) ....................................................................................
1
= 12 x 2 +16 x−3 .............................................................................................................
1
b. ( 4 a+b )3 −( 48 a4 b3 c ∶ ( 8 a b2 ×2 bc ) )
=
( 1 ( 4 a )3 +3 ( 4 a )2 ( b )1+ 3 ( 4 a )1 ( b )2 +1 ( b )3 ) −¿
=
( 1 ( 64 a ) +3 ( 16 a ) ( b ) +3 ( 4 a ) ( b ) + 1 ( b ) ) −¿
=
( 64 a3 +48 a 2 ( b )+ 12a ( b2 ) +b3 ) −¿
3
2
2
3
(
(
48 a 4 b3 c
8 a b2 ×2 bc
(
48 a 4 b3 c
16 a b3 c
16 a b3 c × 3 a3
3
16 a b c
) ................................
1
) .................................
1
) .............................................
1
= 64 a 3+ 48 a2 b+12 a b2 +b 3−3 a3 ............................................................................
1
= ( 64 a3−3 a3 ) + 48 a2 b+12 a b2 +b 3 .........................................................................
1
= ( 64−3 ) a3 + 48 a2 b +12 a b2 +b3 .............................................................................
1
= 61 a3 +48 a 2 b+ 12a b 2+ b3 .......................................................................................
1
Skor maksimum:
14
INSTRUMEN TES
NILAI
LEMBAR SOAL TES
Nama Sekolah :
Nama Siswa
Hari/ Tanggal :
Kelas/ No. Absen :
Mata Pelajaran : Matematika
Alokasi Waktu
Kerjakan soal di bawah ini dengan benar dan jelas!
1. Selesaikan operasi penjumlahan berikut!
a.
( x 2+ 3 y ) + (−7 y + 4 x 2−5 )
( x+ 2 y −z )+ (2 x− y +3 z )
b.
2. Selesaikan operasi pengurangan berikut!
7 p−8−q−5 p−2 q−3
a.
3
3
b.
3 m n−8 m−7 n−12−m n−27 m−5 n
3. Selesaikan operasi kali berikut!
( 3 x−5 ) ( 2 x +6 )
a.
b.
( 4 x2 − y ) ( x +2 x−7 )
4. Selesaikan operasi bagi berikut!
3 2
a.
64 a b ∶ 4 ab
b.
28 a4 b c3 ∶ ( 2 abc × 7 a2 c )
5. Selesaikan operasi pangkat berikut!
a.
( 4 a+7 b )2
b.
( 9 p−5 p )3
6. Selesaikan operasi berikut!
a.
( 4 x +1 )2− (2 x−2 )2
3
b.
( 4 a+b ) −( 48 a4 b3 c ∶ ( 8 a b2 ×2 bc ) )
:
: 2 x 45 menit
KUNCI JAWABAN TES
1. a. ( x 2+ 3 y ) + (−7 y + 4 x 2−5 )
skor
= ( x 2+ 4 x 2 ) + ( 3 y−7 y ) −5 ....................................
1
= ( 1+4 ) x 2 + ( 3−7 ) y−5 ........................................
1
= 5 x2 −4 y −5 .............................................................
1
b. ( x+ 2 y −z )+ (2 x− y +3 z )
= ( x+ 2 x ) + ( 2 y− y ) + (−z+3 z ) .............................
1
= ( 1+2 ) x + ( 2−1 ) y + (−1+3 ) z .............................
1
= 3 x+ y +2 z .................................................................
1
Skor maksimum:
6
2. a. 7 p−8−q−5 p−2 q−3
= ( 7 p−5 p ) + (−q−2 q ) + (−8−3 ) ...............................
1
= ( 7−5 ) p+ (−1−2 ) q+ (−11 ) ......................................
1
=
2 p−3 q−11 ....................................................................
1
b. 3 m3 n−8 m−7 n−12−m3 n−27 m−5 n
= ( 3 m3 n−m3 n ) + (−8 m−27 m )+ (−7 n−5 n )−12 ...............................................
1
= ( 3−1 ) m3 n+ (−8−27 ) m+ (−7−5 ) n−12 ..........................................................
1
= 2 m3 n−35 m−12n−12 ............................................................................................ 1
Skor maksimum:
3. a. ( 3 x−5 ) ( 2 x +6 )
6
= 3 x ( 2 x+6 )−5 ( 2 x+6 ) ................................................................
1
= 3 x ( 2 x )+3 x ( 6 )−5 ( 2 x )−5 ( 6 ) ...................................................
1
= 6 x 2+18 x−10 x−30 .................................................................. 1
= 6 x 2+ (18 x−10 x )−30 ...............................................................
1
= 6 x 2+ (18−10 ) x−30 .................................................................
1
= 6 x 2+ 8 x−30 ................................................................................
1
b. ( 4 x2 − y ) ( x +2 x−7 ) = 4 x 2 ( x+2 x−7 )− y ( x+2 x−7 ) .........................................
1
= 4 x 2 ( x )+ 4 x 2 ( 2 x )+ 4 x 2 (−7 )− y ( x ) − y ( 2 x )− y (−7 ) ....
1
= 4 x 3 +8 x 3−28 x 2−xy−2 xy +7 y .....................................
1
= ( 4 x3 +8 x 3 )−28 x 2 + (−xy−2 xy )+ 7 y ...........................
1
= ( 4 +8 ) x 3−28 x 2+ (−1−2 ) xy +7 y ..................................
1
= 12 x 3−28 x 2−3 xy +7 y ....................................................... 1
4.
Skor maksimum
12
3 2
a. 64 a b ∶ 4 ab
=
3
2
64 a b
4 ab
..................................................................................................
1
2
4 ab × 16 a b
4 ab
=
...........................................................................................
=
1
2
16 a b .....................................................................................................
1
b. 28 a4 b c3 ∶ ( 2 abc × 7 a2 c )
=
28 a4 b c 3
2 abc ×7 a 2 c
........................................................................
=
............................................................................
1
28 a4 b c 3
14 a3 b c 2
1
3
2
14 a b c × 2ac
3
2
14 a b c
=
...................................................................
1
2 ac
=
....................................................................................
1
Skor maksimum:
7
5. a. ( 4 a+7 b )2
2
=
1
1
2
..............................................................
1
= 1 ( 16 a2 ) +2 ( 4 a ) (7 b ) +1 ( 49 b2 ) .............................................................
1
1 ( 4 a ) +2 ( 4 a ) ( 7 b ) +1 (7 b )
= 16 a2 +8 a ( 7 b )+ 49 b2 .............................................................................. 1
= 16 a2 +56 ab+ 49 b2 ................................................................................. 1
b. ( 9 p−5 p )3
3
=
2
1
1
2
1 ( 9 p ) +3 ( 9 p ) (−5 p ) +3 ( 9 p ) (−5 p ) + 1 (−5 p )
3
...........................
1
= 1 ( 729 p3 ) +3 ( 81 p2 ) (−5 p ) +3 ( 9 p ) ( 25 p 2 ) +1 ( −125 p3 ) ....................
1
= 729 p 3+243 p2 (−5 p ) +27 p ( 25 p2 )−125 p3 .........................................
1
= 729 p 3−1215 p 3+ 675 p3 −125 p3 .........................................................
1
= 64 p3 ............................................................................................................
1
Skor maksimum
9
6. a. ( 4 x +1 )2− (2 x−2 )2
=
( 1 ( 4 x )2 +2 ( 4 x )1 (1 )1 +1 ( 1 )2 )−( 1 ( 2 x )2+2 ( 2 x )1 (−2 )1 +1 (−2 )2)
=
( 1 ( 16 x 2) + 2 ( 4 x )( 1 ) +1 ( 1 ) )−( 1 ( 4 x 2 ) +2 ( 2 x ) (−2 ) +1 ( 4 ) )
=
( 16 x 2+ 8 x ( 1 ) +1 )−( 4 x 2 +4 x (−2 )+ 4 ) ...............................................................
..................
.............................
1
1
1
= 16 x 2+8 x +1−4 x 2+8 x−4 ...................................................................................
1
= ( 16 x 2−4 x 2 ) + ( 8 x +8 x ) + ( 1−4 ) ..........................................................................
1
= ( 16−4 ) x 2 + ( 8+8 ) x + (−3 ) ....................................................................................
1
= 12 x 2 +16 x−3 .............................................................................................................
1
b. ( 4 a+b )3 −( 48 a4 b3 c ∶ ( 8 a b2 ×2 bc ) )
=
( 1 ( 4 a )3 +3 ( 4 a )2 ( b )1+ 3 ( 4 a )1 ( b )2 +1 ( b )3 ) −¿
=
( 1 ( 64 a ) +3 ( 16 a ) ( b ) +3 ( 4 a ) ( b ) + 1 ( b ) ) −¿
=
( 64 a3 +48 a 2 ( b )+ 12a ( b2 ) +b3 ) −¿
3
2
2
3
(
(
48 a 4 b3 c
8 a b2 ×2 bc
(
48 a 4 b3 c
16 a b3 c
16 a b3 c × 3 a3
3
16 a b c
) ................................
1
) .................................
1
) .............................................
1
= 64 a 3+ 48 a2 b+12 a b2 +b 3−3 a3 ............................................................................
1
= ( 64 a3−3 a3 ) + 48 a2 b+12 a b2 +b 3 .........................................................................
1
= ( 64−3 ) a3 + 48 a2 b +12 a b2 +b3 .............................................................................
1
= 61 a3 +48 a 2 b+ 12a b 2+ b3 .......................................................................................
1
Skor maksimum:
14