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Matematik Tingkatan 2 Bab 2

PBD
PBD 2.3 Ungkapan Algebra dan Hukum Operasi Asas Aritmetik Buku Teks
PBD
Algebraic Expressions and Laws of Basic Arithmetic Operations ms. 34 – 38
13. Permudahkan setiap yang berikut. SP 2.3.1 TP3
Simplify each of the following.
(a) (h + k)(h – k) – (h + k ) (b) (p + q) + (2p + 3q)(3p – 2q)
2
2
2
(i) Ungkapan algebra = h – k – h – k 2 = p + 2pq + q + 6p – 4pq
2
2
2
2
2
2
Algebraic expressions = –2k + 9pq – 6q 2
2
3x(x + y) + 5y(x – 2y) = 7p + 7pq – 5q 2
2
= 3x + 3xy + 5xy – 10y 2
2
= 3x + 8xy – 10y 2
2
(ii) Penyebut yang sama 3p + 2q p – 5q 4m – 3n 3m – 4n
Same denominators (c) p – 2q – p – 2q (d) 2m + 3n + 2m + 3n
z – 5z = z – 5z 3p + 2q – p + 5q 4m – 3n + 3m + 4n
2y 2y 2y = p – 2q = 2m + 3n
–4z
= 2p + 7q 7m + n
2y = p – 2q = 2m + 3n
2z
= –
y

(iii) Penyebut dengan gandaan (e) 3n + 5n (f) 3h 2 – 7h 2
bagi penyebut lain 12m 2 4m 2 2k 10k
Denominators with the multiple of 3n 5n × 3 3h × 5 7h 2
2
the other denominator = 12m 2 + 4m × 3 = 2k × 5 – 10k
2
5 – 3 2 2
4c 2c = 3n + 15n = 15h – 7h
= 5 – 3 × 2 Samakan 12m 2 10k
4c 2c × 2 penyebut. = 18n = 8h 2
Equalise the
= 5 – 6 = – 1 denominators. 12m 2 10k
4c 4c 4h 2
=
5k
(iv) Penyebut tanpa faktor (g) c – 3 (h) 1 + 5
sepunya 5d 4c 7z 6z
Denominators without common c × 4c 3 × 5d 1 × 6 5 × 7
factor = 5d × 4c – 4c × 5d = 7z × 6 + 6z × 7
3 5
2
+ 4c – 15d 6 + 35
4u 3v = 20cd = 42z
3 × 3v 5 × 4u
= + 41
4u × 3v 3v × 4u = 42z
= 9v + 20u
12uv

(v) Penyebut dengan faktor (i) p – 3p (j) 3 + n
sepunya 6q 10qr 2mn 6m 2
Denominators with common factor = p × 5r – 3p × 3 = 3 × 3m + n × n
2
3y + 5z 6q × 5r 10qr × 3 2mn × 3m 6m × n
4x 6x = 5pr – 9p = 9m + n 2
2
= 3y × 3 + 5z × 2 2x 4x, 6x 30qr 6m n
2, 3
4x × 3 6x × 2 GSTK/ HCF
= 9y + 10z = 2x × 2 × 3
12x = 12x

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