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balenkosanya
@balenkosanya
August 2022
1
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Упростить(n-1)!/(n+2)!
Вычислить P6-P4/P4
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moboqe
(n-1)!/(n+2)!=(1*2*3*...*(n-1))/(1*2*3...*(n-1)*n*(n+1)*(n+2))=1/(n*(n+1)*(n+2))=1/((n^2+n)*(n+2))=1/(n^3+3n^2+2n)
P6=6!
P4=4!
(6!-4!)/4!=(4!(5*6-1))/4!=29
2 votes
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Answers & Comments
P6=6!
P4=4!
(6!-4!)/4!=(4!(5*6-1))/4!=29