Question

Ajutati sa rezolv va rog mult ))

Answer 1

An,4-aranjamente de n luate cate 4 si asa si la celelalte
[n!/(n-4)! +n!/(n-3)!]/(n-2).n!/(n-3)!=
se da factor si se simplifica cu n!  apoi
se da factor la numarator 
1/(n-4)![1+1/(n-3)]/(n-2)/(n-3)(n-4)!
simplificand
[(n-3+1)/(n-3)]/(n-2)(n-3)=1

k)
5!/(5!/2!.3! +6!/4!)=
se da 5! factor si se simplifica
1/(1/2.2.3+6/4.3.2)=1/(3/24)=24/3=8

An,2=12
n!/(n-2)!=12
n(n-1)(n-2)!=12(n-2)! |:(n-2)!
n(n-1)=12
n²-n-12=0
n1,2=(1+,-√(1+48))/2⇒n1=5 si n2=-4
 
An,2=6
n!/(n-2)!=6
n!=6(n-2)!
n(n-1)(n-2)!=6(n-2)!
n²-n-6=0
n1=3,n2=-2

P(n+3)=720.An,5.P(n-5)
(n+3)!=720.n!/(n-5)!  .(n-5)! se simplifica (n-5)!
(n+3)(n+2)(n+1)n!=720.n!
(n+3)(n+2)(n+1)=720
aici ai de rezolvat ec de gr 3