Page 32 - Hybrid PBD Matematik TG3_Edisi Guru

Matematik Tingkatan 3 Bab 1
27. Cari nilai bagi setiap yang berikut. SP 1.2.5 TP3
Find the value of each of the following.
1 4 1 5 3
5
2
4
1 (a) 16 = √16 (b) 243 = √243 (c) 9 = (√9 ) 3
3
3
27 = √27 = 2 = 3 = 3 3
= 3 = 27
3 5 2 2
4
3
5
6
4
5
3
6
3
5
2
(d) 81 = ( √81) (e) 729 = (√729 ) (f) 125 = ( √125) (g) 32 = (√32 )
2
= 3 3 = 3 5 = (5) 2 = (2) 2
= 27 = 243 = 25 = 4
28. Padankan setiap berikut dengan nilai yang betul. SP 1.2.5 TP3
Match each of the following with the correct value.
1
2 ÷ 2 2 2 × 2 2 2 –3 2 0 2 2 (2 )
3
3
3 2
1 1 √2 2 32 64
8

29. Permudahkan setiap yang berikut. SP 1.2.6 TP4
Simplify each of the following.

1 3 3 1
2 4 2
12 4 4
4
–4 5
2 2 3
1 2 (3x y ) × 3 x y 8 p q × (8p q )
(2m n ) × 2 m n (a) 18x y (b) 1
9 6 3
3
–2 –4
2 –3
3 10 4 2
m n 1 1 1 3 (2 p q )
–3 5
3
1
1
1
2
2
2
y
2 2 3
1
1 9( ) 6( ) 2 3 x 2( ) 4( ) × 3 x y 3( ) 3( ) 12( ) 4( )
1
–4 5
2 m 3 n 3 × 2 m n = 2 –3 = 2 4 p q × 2 4 p 4 q 4
3
3
–2 –4
= 18x y 1 1 1
m n 1 3 2 3( ) 10( ) 4( )
–3 5
2
q
2
2
p
1 + 2 3 2 + 2 9 3
+
y
n
= 2 3 3 m 3 + (–2) – (–3) 2 + (–4) – 5 = 1 18 2 x 1 + 2 – 2 2 + 3 – (–3) 1 2 4 4 –4 + 3 – 5 5 + 1 – 2
3 2
= 2m n 3 2 = 2 2 p q
4 –7
8
= 2m 4 = 1 2 xy 3
18
n 7 1 = 2 p q
2
–6 4
= xy
8
3
–6 4
2 = √2 p q
30. Hitung nilai bagi setiap yang berikut. SP 1.2.6 TP4
Calculate the value of each of the following.
1 1
3
4
4 3
2 12
(25 ) × (25 ) ( √16) × 5 –2
1
1 3( )(4) (a) (b) 1
–2 2
(8 ) × 3 –2 = 2 3 × 3 –2 (2 ) (16 )
3 4
5 4
1 3(2)( ) 2(4)( ) 2(2)( 1 )
1
1
1
2 3
(8 ) 2 3 = 5 3 × 5 12 2 4( )(3) × 5 –2
4
= 2 4 – 2 × 3 –2 8 2 –2(2) = 4(5)( )
1
4
1
3
= 2 2 5 × 5 3 = 2 3 – 5 2 –2
× 5
3 2 = 2 –4 1
= 4 8 + 1 = 2 × 5 2
2
9 = 5 3 3 × 2
4
= 5 × 2 4 = 1
3
= 2 000 100
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