Page 99 - Matematik Tambahan Tingkatan 4 KSSM

Indeks, Surd dan Logaritma

Contoh 1

Permudahkan ungkapan algebra yang berikut.
2n
4 × 4 m 3 m + 2 – 3 m
(a) (b)
4 n 3 m
2
3 2
−1 3
3 – 4
− 4
(c) (5x ) × 4xy ÷ (xy) (d) 4a b × (4ab )
Penyelesaian
m
2
2n
4 × 4 m 2n + m – n 3 m + 2 – 3 m 3 × 3 – 3 m
(a) = 4 (b) =
4 n 3 m 3 m
m
2
= 4 n + m 3 (3 – 1)
=
3 m
= 8
3 2
3 – 4
2
–1 3
(c) (5x ) × 4xy ÷ (xy) – 4 (d) 4a b × (4ab )
–1 3
(5x ) × 4xy 2 1
3 2
= = 4a b ×
3 4
(xy) – 4 (4ab )
–3
2
3
3 2
= 5 x × 4xy × (xy) 4 4a b BAB 4
=
4 12
= 125 × 4 × x –3 + 1 + 4 × y 2 + 4 256a b

2 6
= 500x y 1
= 10
64ab
Contoh 2
Permudahkan ungkapan algebra yang berikut.
1 –1 –2
– 2a
(a) a × 2a 2 (b)
3
– 3
a 2
1 3 1 – 1

–
2
3
2
–3
2
(c) a × a (d) a (a 2 + 2a 2 – 3a )
2
Penyelesaian
1 1 1 1 –2 3
– – – – 2a –2 –
(a) a × 2a = 2 × a × a 2 (b) = 2a ÷ a
2
3
2
3
– 3
a 2
1
1
– + ( – ) –2 – ( – )
3
= 2a 3 2 = 2a 2
5 1
– –
6
= 2a = 2a 2
2 2
= =
5 1
a 6 a 2
2 – 3 1 3 1 – 1

–
–3
2
2
2
3
(c) a × a = a × a (d) a (a + 2a – 3a )
2
2
2
3
2
1
2
3
1
3 ( – ) – 1 3 – 1 – –
1
+
2
2
= a 2 = a × a + a × 2a – a × 3a 2
2
2
2
1
1
2 – 3 – + 3 – + 1 – – 1
1
= a 3 2 = a 2 2 + 2a 2 2 – 3a 2 2
5
0
1
= a – = a + 2a – 3a –1
6
1 3
= = a + 2 –
5 a
a 6
4.1.1 91