3.2.1 Ungkapan Algebra II, Praktis Berformat PT3


3.2.1 Ungkapan Algebra II, Praktis Berformat PT3
 
Soalan 1:
Hitungkan hasil darab bagi setiap sebutan algebra yang berikut.
(a) 2rs × 4r2s3t
(b) 1 2 5 a 2 b 2 × 5 14 a b 3 c 2 (c) 1 1 2 xy× 4 9 x 2 z  

Penyelesaian:
(a)
2rs × 4r2s3t = 2 × r × × 4 × r × r × s × s × s × t = 8 r3s4t

(b) 1 2 5 a 2 b 2 × 5 14 a b 3 c 2 = 7 1 5 1 aabb× 5 1 14 2 abbbcc = 1 2 a 3 b 5 c 2

(c) 1 1 2 xy× 4 9 x 2 z = 3 1 2 1 xy× 4 2 9 3 xxz = 2 3 x 3 yz


Soalan 2:
Cari hasil bahagi bagi setiap sebutan algebra yang berikut.
(a)  48 a 2 b 3 c 4 16a b 2 c 2  
(b) 15e f 3 g 2 ÷( 40e f 2 g ) (c) 5 a 3 c 2 ÷ 1 2 ac  

Penyelesaian:
(a)  48 a 2 b 3 c 4 16a b 2 c 2 = 48 3 a ×a× b × b ×b× c × c ×c×c 16 1 a × b × b × c × c =3a c 2
(b) 15e f 3 g 2 ÷( 40e f 2 g ) = 15 3 e × f × f ×f× g ×g 40 8 e × f × f × g = 3 8 fg

(c) 5 a 3 c 2 ÷ 1 2 ac=5a 3 2 c 2 × 2 a c =10 a 2 c


Soalan 3:
(– 4a2b) ÷ 3b2c× 9ac =

Penyelesaian:
( 4 a 2 b )÷3 b 2 c 2 ×9ac = 4 a 2 b 3 b 2 1 c 2 1 × 9 3 a c = 12 a 3 bc



Soalan 4:
2a2b × 14b3c ÷ 56ab2c2 =

Penyelesaian:
2 a 2 b×14 b 3 c÷56a b 2 c 2 = 2 a 2 1 b× 14 b 3 1 c 56 2 a b 2 c 2 1 = a b 2 2c


Soalan 5:
2 3 ( 3a6b+ 3 4 c )=  

Penyelesaian:
2 3 ( 3a6b+ 3 4 c ) = 2 3 × 3 a 2 3 ( 6 2 b ) 2 3 ( 3 4 2 c ) =2a+4b 1 2 c

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