Question

1.) Tentukan hasil perpangkatan bentuk aljabar berikut.

A.) (5a)³ = ....
B.) (2xy)² = ....
C.) (-3x)³ = ....
D.) (4p²q)² = ....
E.) (-5xy³)⁴ = ....
F.) -(2abc)³ = ....
G.) -(3pq)⁴ = ....
H.) a(ab²)³ = ....

2.) Jabbarkan perpangkatan bentuk aljabar berikut.

A.) (x + 4)³ = ....
B.) (a - 5)⁴ = ....
C.) (2x + y)³ = ....
D.) (3p + q)⁴ = ....
E.) (3m - 2n) = ....
F.) (4a - 3b)³ = ....
G.) (2y² + y)³ = ....
H.) (3b - 2)^5 = ....


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"mohon bantuan"

Answer 1

a. (5a)³ 
= (5a) x (5a) x (5a)
= (25a²) x (5a) 
= 125a³

b. (2xy)²
= (2xy) x (2xy) 
= (4xy²) 

c. (-3x)³
= (-3x) x (-3x) x (-3x)
= (9x²) x (-3x)
= -27x³

d. (4p²q)²
= (4p²q) x (4p²q)
= 16p⁴q²

e.(-5xy³)⁴ = ....
= (-5)⁴ x⁴ ypangkat 3 x 4
= 625 x⁴y¹²

f.(5xy) ⁴
= (5x5x5x5)(x.x.x.x)(y.y.y.y)
= 625 .x⁴ . y⁴
= 625 x⁴y⁴

g. -(3pq)⁴ = ....
=(-3 x -3 x-3 x -3)(p.p.p.p)(q.q.q.q)
= 81p⁴q⁴

h. a(ab²)³ = ....
= a(a³b⁶)
= a⁴b⁶

2)
a.(x + 4)³ = ....
= x³ - 64

b.(a - 5)⁴ = ....
= (a)² - 2.(a).5 + 5²
= 2a² - 5a + 25

c. (2x + y)³ = ...
=(2x)³ + 3(2x)²y + 3(2x)y² + y³
= 8x³ + 12x²y + 6xy² + y³

d.(3p + q)⁴ = ....
= (3p + q) (3p + q) (3p + q) (3p + q)
= (9p² + 3pq + 3pq + q²) (9p² + 3pq + 3pq + q²)
= (9p² + 6pq + q²) (9p² + 6pq + q²)
= (81p⁴ + 54p³q + 9p²q² + 54p³q + 36p²q² + 6pq³ + 9p²q² + 6pq³ + q⁴)
= 81p⁴ + 108p³q + 54p²q² + 12pq³ + q⁴

e. (3m - 2n) = ....
=?


f.(4a - 3b)³ =..
= ( 4a - 3) (4a - 3) (4a - 3)
= (16a² -12a - 12a + 9) (4a-3)
= (16a² - 24a + 9) (4a-3)
=(64a³-48a²-96a²+72a+36a-27)
= (64a³ - 144a²+108a - 27)

g.(2x+y)³ =
 =(2x + y) (2x + y) (2x + y)
= (4x² + 2xy + 2xy + y²) (2x + y)
= (4x² + 4xy + y²) (2x + y)
= 8x³ + 8x²y + 2xy² + 4x²y + 4xy² +y³
= 8x³ + y³ + 6xy² + 12x²y

maaf 2 lagi belum di kerjain ,,,,,,,,