Question

Dau coroana !!!Va rog daca stiti ajutati-ma , as vrea pas cu pas rezolvarea , daca stiti si vreti pls help me!
Sa se rezolve ecuatiile :
a) (n+2)! / n! = 12 , n € N ;
b) (n+1)! / 12 = (n-1)! , n € N ;
c) (n-3)! / (n-5)! = 6 , n € N ;

Answer 1

n!=1·2·3·.....·n (n factorial)
a. (n+2)!=n!·(n+1)(n+2)
(n+1)(n+2)=12
n²+n+2n+2=12
n²+3n-10=0
n1,2=[-3+/-√(9+40)]/2=(-3+/-7)/2⇒n∈{-5, 2}∩N={2}
b. (n-1)!·n(n+1)/12=(n-1)!
n(n+1)/12=1
n²+n=12
n²+n-12=0
n1,2=[-1+/-√(1+48)]/2=(-1+/-7)/2⇒n∈{-4, 3}∩N={3}
c. (n-5)!(n-4)(n-3)/(n-5)!=6
(n-4)(n-3)=6
n²-3n-4n+12=6
n²-7n+6=0
n1,2=[7+/-√(49-24)]/2=(7+/-5)/2⇒n∈{1, 6}∩N={1,6}