suku aljabar\lampiran-lampiran\Lampiran 6 INSTRUMEN SOAL TES(aljabar)

INSTRUMEN TES LEMBAR SOAL TES

Nama Sekolah : Nama Siswa :

Hari/ Tanggal : Kelas/ No. Absen :

Mata Pelajaran : Matematika Alokasi Waktu : 2 x 45 menit

Kerjakan soal di bawah ini dengan benar dan jelas! 1. Selesaikan operasi penjumlahan berikut!

a.

(

x2+3y

)

+

(

−7y+4x2−5) b. (x+2y−z)+ (2x−y+3z)

2. Selesaikan operasi pengurangan berikut! a. 7p−8−q−5p−2q−3

b. 3m3_n

−8m−7n−12−m3_n

−27m−5n 3. Selesaikan operasi kali berikut!

a. (3x−5) (2x+6) b.

(

4x2

−y

)

(x+2x−7) 4. Selesaikan operasi bagi berikut!

a. 64a3b2∶4ab b. 28a4_{b c}3

∶

(

2abc ×7a2_c

)

5. Selesaikan operasi pangkat berikut!

a. (4a+7b)2 b. (9p−5p)3

6. Selesaikan operasi berikut! a. (4x+1)2−(2x−2)2 b. (4a+b)3−

(

48a4_b3_c

∶

(

8a b2_×2bc

)

)

Lampiran 6

KUNCI JAWABAN TES skor 1. a.

(

x2+3y

)

+

(

−7y+4x2−5) =

(

x2+4x2

)

+(3y−7y)−5 ... 1

= (1+4)x2

+(3−7)y−5 ... 1 = 5x2−4y−5 ... 1

b. (x+2y−z)+ (2x−y+3z) = (x+2x)+(2y−y)+(−z+3z) ... 1 = (1+2)x+(2−1)y+(−1+3)z ... 1 = 3x+y+2z ... 1

Skor maksimum: 6

2. a. 7p−8−q−5p−2q−3 = (7p−5p)+(−q−2q)+(−8−3) ... 1 = (7−5)p+(−1−2)q+(−11) ... 1

= 2p−3q−11 ... 1

b. 3m3_n

−8m−7n−12−m3_n

−27m−5n

=

(3

m3n−m3n

)

+(−8m−27m)+(−7n−5n)−12 ... 1 = (3−1)m3n+(−8−27)m+(−7−5)n−12 ... 1 = 2m3_n

−35m−12n−12 ... 1

Skor maksimum: 6

3. a. (3x−5) (2x+6) = 3x(2x+6)−5(2x+6) ... 1 = 3x(2x)+3x(6)−5(2x)−5(6) ... 1 = 6x2

+18x−10x−30 ... 1

= 6x2

= 6x2

+ (18−10)x−30 ... 1 = 6x2+8x−30 ...

1

b.

(

4x2−y

)

(x+2x−7) = 4x2(x+2x−7)−y(x+2x−7) ... 1 = 4x2(x)+4x2(2x)+4x2(−7)−y(x)−y(2x)−y(−7) .... 1 = 4x3+8x3−28x2−xy−2xy+7y ... 1

=

(4

x3 +8x3

)

−28x2

+(−xy−2xy)+7y ... 1 = (4+8)x3−28x2+(−1−2)xy+7y ... 1 = 12x3−28x2−3xy+7y ... 1

Skor maksimum 12

4.

a. 64a3_b2_{∶4ab =}

64a3_b2

4ab

...

1

= 4ab ×16a2b

4ab

... 1

= 16a2_{b ...} 1

b. 28a4_{b c}3

∶

(

2abc ×7a2_c

)

₌ 28a

4

b c3

2abc ×7a2c

... 1

= 28a

= 14a3b c2×2ac 14a3b c2

... 1

= 2ac

... 1

Skor maksimum: 7

5. a. (4a+7b)2 =

1(4a)2+2(4a)1(7b)1+1(7b)2 ... 1 = 1(16a2

)

+2(4a) (7b)+1(49b2

)

_{... 1} = 16a2+8a(7b)+49b2 ... 1 = 16a2

+56ab+49b2 _{... 1}

b. (9p−5p)3 =

1(9p)3+3(9p)2(−5p)1+3(9p)1(−5p)2+1(−5p)3 ... 1

= 1(729p3

)

+3

(81

p2

)

(−5p)+3(9p)

(

25p2

)

+1(−125p3

)

... 1

= 729p3

+243p2_(−5p)+27p

(25

_p2

)

−125p3 _{... 1} = 729p3−1215p3+675p3−125p3 ... 1 = 64p3 ...

1

Skor maksimum 9

6. a. (4x+1)2−(2x−2)2

=

(1

(4x)2+2(4x)1(1)1+1(1)2

)

−

(1

(2x)2+2(2x)1(−2)1+1(−2)2

)

... 1

=

(

1

(16

x2

)

+2(4x)(1)+1(1)

)

−

(

1(4x2

)

+2(2x) (−2)+1(4)

)

... 1

= 16x2+8x+1−4x2+8x−4 ... 1 =

(16

x2

−4x2

)

+(8x+8x)+(1−4) ... 1 = (16−4)x2+(8+8)x+(−3) ... 1 = 12x2+16x−3 ...

1

b. (4a+b)3−

(

48a4b3c∶

(

8a b2×2bc

)

)

=

(1

(4a)3+3(4a)2(b)1+3(4a)1(b)2+1(b)3

)

−¿

(

48a

4_b3_c

8a b2_×2bc

)

... 1

=

(

1

(64

a3

)

+3

(

16a2

)

(b)+3(4a)

(

b2

)

+1(b3

)

)

−¿

(

48a

4

b3c

16a b3c

)

... 1

=

(

64a3+48a2(b)+12a

(

b2

)

+b3

)

−¿

(

16a b

3

c ×3a3

16a b3c

)

...

1

= 64a3

+48a2_b

+12a b2 +b3

−3a3 _{... 1} =

(64

a3

−3a3

)

+48a2_b

+12a b2

+b3 _{... 1}

= (64−3)a3+48a2b+12a b2+b3 ... 1 = 61a3

+48a2_b

+12a b2

+b3 _{... 1}