Matematika

Pertanyaan

sederhanakan bentuk bilangan berpangkat ini!

1.) (2x³y²) (x²y)²

2.)
(A.)3 pangkat m+n × 3 pangkat n-1 × 3 pangkat n-m
-------------------------------------------------(Per)
3 pangkat 2n+1 × 3 pangkat n+1
(B.)(2 pangkat a+2)² - 2²×2 pangkat 2a
---------------------------------------------(per)
2panvkat a × 2 pangkat a+2

1 Jawaban

  • 1)
    (2x^3y^2)(x^2y)^2
    = (2x^3y^2)(x^4y^2)
    = 2x^7y^4

    2a)
    [(3^(m+n) × 3^(n-1) × 3^(n-m)]/[3^(2n+1) × 3^(n+1)]
    = [3^(m+n+n-1+n-m)]/[3^(2n+1+n+1)]
    = [3^(3n-1)]/[3^(3n+2)]
    = 3^(3n-1-(3n+2))
    = 3^(3n-1-3n-2)
    = 3^(-3)
    = 1/3^3
    = 1/(3×3×3)
    = 1/27

    b)
    [(2^(a+2))^2 - 2^2 × 2^(2a)]/[2^a × 2^(a+2)]
    = [(2^(2a+4) - 2^(2a+2)]/[2^(2a+2)]
    = [(2^(2a+4)/2^(2a+2)] - [2^(2a+2)/2^(2a+2)]
    = 2^2 - 2^0
    = 4 - 1
    = 3

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