Сократите дробь (18^(n+1))/((3^(2n+5))*(2^(n-2))). Created by belovaelenaig. matematika-ru

((3^(2n+5))*(2^(n-2)))...

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Сократите дробь
(18^(n+1))/((3^(2n+5))*(2^(n-2)))

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ramit08 (18^(n+1))/((3^(2n+5))*(2^(n-2)))=(2^(n+1))(9^(n+1))/((3^(2n+5))*(2^(n-2)))=(2^(n+1))(3^(2n+2))/((3^(2n+5))*(2^(n-2)))=(2^(n+1-n+2))(3^(2n+2-2n-5))=2^3*3^-3=8/27

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