Question

Sa se rezolve urmatoarele ecuatii :
a) ( n + 6 ) ! = 8 ( n + 5 ) ! ;
b) 6 ( n + 2 ) ! = ( n + 3 ) ! ;
c) ( n + 3 ) ! - ( n + 2 ) ! = ( n + 1 ) ! ;
d) ( 2n + 1 ) ! = 20 ( 2n - 1 ) ! . Va rog frumos

Answer 1

a: (n+6)!=8(n+5)! => (n+6)(n+5)!=8(n+5)! => n+6=8 => n=2
b: 6(n+2)!=(n+3)! => 6=n+2 => 4=n
c: (n+3)!-(n+2)!=(n+1)! => (n+2)!(n+3-1)=(n+1)! => (n+1)!(n+2)(n+2)=(n+1)! => n^2+4n+4=1 => dupa delta ..... n=-2
d: (2n+1)!=20(2n-1)! => 2n(2n+1)=20 => 2n^2+2=10 => iar delta ... n=2