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16^2m ÷ 8^2n ÷ 4^m ×(125/4)^m-n
人气:273 ℃ 时间:2020-06-09 17:10:53
解答
16^2m ÷ 8^2n ÷ 4^m ×(125/4)^m-n
=(2^4)^2m÷(2^3)^2n ÷(2^2)^m×(5^3/2^2)^m-n
=2^8m÷2^6n ÷2^2m×5^(3m-3n)÷2^(2m-2n)
=2^[8m-6n-2m-(2m-2n)] ×5^(3m-3n)
=2^(4m-4n)×5^(3m-3n)
=(2^4)^(m-n)×(5^3)^(m-n)
=16^(m-n)×125^(m-n)
=(16×125)^(m-n)
=2000^(m-n)
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