(56 * 98^(n+2)) / (2^(n+3)*7^(2*n+5))
Сократить дробь....объясните, пожалуйста.
98^(n+2) = (2*49)^(n+2) = 2^(n+2) * 7^(2n+4)
56*98^(n+2) 56 * 2^(n+2) * 7^(2n+4)
---------------------- = ----------------------------------
2^(n+3) * 7^(2n+5) 2^(n+3) * 7^(2n+5)
56 * 2^(n+2-n-3) * 7^(2n+4-2n-5) = 56 * 2^(-1) * 7^(-1) = 56/(2*7) = 4
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Verified answer
98^(n+2) = (2*49)^(n+2) = 2^(n+2) * 7^(2n+4)
56*98^(n+2) 56 * 2^(n+2) * 7^(2n+4)
---------------------- = ----------------------------------
2^(n+3) * 7^(2n+5) 2^(n+3) * 7^(2n+5)
56 * 2^(n+2-n-3) * 7^(2n+4-2n-5) = 56 * 2^(-1) * 7^(-1) = 56/(2*7) = 4