Question

1.tentukan hasil perpangkatan aljabar berikut
A.(2a)²
B.(3xy)³
C.(-2ab)pangkat 4
D.(4a²b²)²
E,-3(x²y)³
F.-(2pq)pangkat 4
G.0,5 (2xy)²
H.a(ab²)³

Answer 1

A) (2a)² = 4a²
b) (3xy)³ = 27xy³
c) (-2ab)^4 = 16ab^4
d) (4a²b²)² = 16a^4b^4
e) -3 (x²y)³ = -3 (x^6 y³) = -3x^6 y³
f) -(2pq)^4 = -16pq^4
g) 0,5 (2xy)² = 0,5 × 4xy² = 2xy²
h) a (ab²)³ = a (a³b^6) = a^4b^6

Answer 2

A. (2a)² = (2a) (2a) → (2a)² = 4
B. (3xy)³= (3xy) (3xy) (3xy) → (3xy)³ = 9x³y³
C. (–2ab)⁴ = (–2ab) (–2ab) (–2ab) (–2ab) → (–2ab)⁴ = 8⁴4b⁴
D. (4a²b²)² = (4a²b²) (4a²b²) → (4a²b²)² = 16a⁴b⁴
E.–3(x²y)³ = –3(x²y) (x²y) (x²y) → –3(x²y)³ = –3 (x6y3
F  –(2pq)⁴ = –(2pq) (2pq) (2pq) (2pq) → –(2pq)⁴ = –16p⁴q⁴
G  ½(2xy)² = ½(2xy) (2xy) →  ½(2xy)2 = ½.4x²y² → ½(2xy)² = 2x²y²
H  a(ab²)³ = a(ab²) (ab²) (ab²) → a(ab²)³ = a(a³b⁶) → a(ab²)³ = a⁴b⁶