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Kuukki
@Kuukki
September 2021
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Sos!
найдите значение выражения , если n натуральное число:
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Universalka
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1)4^(n+1) /2^(2n-1) = 2^(2n+2) / 2^(2n-1) = 2^(2n+2-2n+1) = 2³ = 8
2) 5^(2n+3) / 25^(n-1) = 5^(2n+3) / 5^(2n-2) = 5^(2n+3-2n+2) = 5^5 = 3125
3) [(-5)^(6n+2) * 5^(-4n)] : 25^n = 5^(6n+2-4n) / 5^2n = 5^(2n+2) / 5^2n =
= 5^(2n+2-2n) = 5² = 25
4) 18^(n+3) / [3^(2n+5) * 2^(n-2)] = [2^(n+3) * 3^(2n+6)] / [3^(2n+5) * 2^(n-2)]=
= 2^(n+3-n+2) * 3^(2n+6-2n-5) = 2^5 * 3 = 32 * 3 = 96
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Verified answer
1)4^(n+1) /2^(2n-1) = 2^(2n+2) / 2^(2n-1) = 2^(2n+2-2n+1) = 2³ = 82) 5^(2n+3) / 25^(n-1) = 5^(2n+3) / 5^(2n-2) = 5^(2n+3-2n+2) = 5^5 = 3125
3) [(-5)^(6n+2) * 5^(-4n)] : 25^n = 5^(6n+2-4n) / 5^2n = 5^(2n+2) / 5^2n =
= 5^(2n+2-2n) = 5² = 25
4) 18^(n+3) / [3^(2n+5) * 2^(n-2)] = [2^(n+3) * 3^(2n+6)] / [3^(2n+5) * 2^(n-2)]=
= 2^(n+3-n+2) * 3^(2n+6-2n-5) = 2^5 * 3 = 32 * 3 = 96