Запишите в виде дроби А)n+2/n!-3n+2/(n+1)! Б)1/(k-1)!-k/(k+1)В)1/(k-2)!-k^3+k/(k+1)!

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Запишите в виде дроби
А)n+2/n!-3n+2/(n+1)!
Б)1/(k-1)!-k/(k+1)
В)1/(k-2)!-k^3+k/(k+1)!


Алгебра (57 баллов) | 35 просмотров
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Может Б)1/(k-1)!-k/(k+1)!?

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Правильный ответ

А
(n+2)/n!-(3n+2)/(n+1)! =(n+2)/n!-(3n+2)/[n!(n+1)]=
=(n²+n+2n+2-3n-2)/[n!(n+1)]=n²/(n+1)!
Б
1/(k-1)!-k/(k+1)!=1/(k-1)!-k/[(k-1)!*k*(n+1)]=
=(k²+k-k)/(k+1)!=k²/(k+1)!
В
1/(k-2)!-(k^3+k)/(k+1)! =1/(k-2)!-(k³+k)/[(k-2)!(k-1)k(k+1)]=
=(k³-k-k³-k)/(k+1)!=-2k/(k+1)!

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I never not boring with my friends

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