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tahirsultanov
@tahirsultanov
July 2022
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найдите модуль разности квадратов двух последовательных нечетных чисел одно из которых составляет 60%другого
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sergey560
| (2n+1)^2 - (2n+3)^2 | = | (2n+1+2n+3)(2n+1-2n-3) | =
| (4n+4)(-2) | = 8(n+1)
2n+1 = 0,6 (2n+3) -> 2n+1 = 1,2n + 1,8 -> 0,8n = 0,8 ->
n = 1 8(n+1) = 8*2 = 16
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Answers & Comments
| (4n+4)(-2) | = 8(n+1)
2n+1 = 0,6 (2n+3) -> 2n+1 = 1,2n + 1,8 -> 0,8n = 0,8 ->
n = 1 8(n+1) = 8*2 = 16