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Sa se rezolve ecuatia: Aranjament de n-1 liate cate 5=18×aranjament de n-3 luate cate 4

Răspuns :

Renatemambouko
Aranjamente de n luate câte m = n(n-1)(n-2)(n-3)...(n-m+1

Aranjamente de n-1 luate cate 5=18×Aranjamente de n-3 luate cate 4
(n-1)(n-1-1)(n-1-2).....(n-1-4)=18(n-3)(n-3-1).....(n-3-3)
(n-1)(n-2)(n-3).....(n-5)=18(n-3)(n-4).....(n-6)
(n-1)(n-2)=18(n-6)
n²-3n+2=18n-108
n²-21n+110=0
Δ=(-21)²-4×110=441-110=1
n₁=(21+1)/2=22/2=11
n₂=(21-1)/2=20/2=10





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