Page 19 - Matematik Tingkata 3

Bab 1 IndeksIndeks
Bab 1

Pembahagian nombor Pendaraban berulang
dalam bentuk indeks
5 faktor 1
5
(a) 4 ÷ 4 2
4 5 4 × 4 × 4 × 4 × 4 = 4 × 4 × 4 = 4 3 BAB
— = —————––––
4 2 4 × 4 3 faktor (Baki)
2 faktor
2
5
4 ÷ 4 = 4 3 3 = 5 – 2
5
2
4 ÷ 4 = 4 5–2
6 faktor
6
(b) 2 ÷ 2 2
2 6 2 × 2 × 2 × 2 × 2 × 2 = 2 × 2 × 2 × 2 = 2 4
— = —————––––—–
2
2 2 × 2 4 faktor (Baki)
2 faktor
6
2
2 ÷ 2 = 2
2 ÷ 2 = 2
2
6
5 faktor
5
(c) (–3) ÷ (–3) 3
(–3) 5 (–3) × (–3) × (–3) × (–3) × (–3) = (–3) × (–3) = (–3) 2
—— = —————––––—–––——––
(–3) × (–3)× (–3)
3

(–3)
3 faktor 2 faktor (Baki)
3
5
(–3) ÷ (–3) = (–3)
3
5
(–3) ÷ (–3) = (–3)
Perbincangan:
Apakah perkaitan antara pembahagian nombor dalam bentuk indeks dengan pendaraban
berulang?
Hasil daripada Cetusan Minda 2, didapati bahawa; BIJAK MINDA
5
4 ÷ 4 = 4 5 – 2 Diberi m a – b = m dan
2
7
2
2 ÷ 2 = 2 6 – 2 0 < a < 10. Jika a > b,
6
3
5
(–3) ÷ (–3) = (–3) 5 – 3 nyatakan nilai-nilai yang
mungkin bagi a dan b.
Secara generalisasi, a ÷ a = a m – n
n
m
Contoh 8
Ringkaskan setiap yang berikut.
4
2
4 3
2
(a) 5 ÷ 5 2 (b) (–3) ÷ (–3) ÷ (–3) (c) m n ÷ m n
4

5
5
(d) 25x y ÷ 5xy (e) 12m ÷ 4m ÷ m 2 (f) –16p ÷ 2p ÷ 4p 2
8
10
2 3
Penyelesaian:
(a) 5 ÷ 5 2 (b) (–3) ÷ (–3) ÷ (–3) (c) m n ÷ m n
4 3
4
2
4
2
2 1
= 5 4 – 2 = (–3) ÷ (–3) ÷ (–3) 1 = m n ÷ m n
2
4

4 3
2
= 5 = (–3) 4 – 2 – 1 = m 4 – 2 3 – 1
n
= (–3) 1 = m n
2 2
= –3 Saiz sebenar
9