có \(4n+13\) chia hết cho \(2n+1\)
=> 4n + 2 +11 chia hết chi 2n+1
=> 2.(2n+1) +11 chia hết cho 2n+1
ta thấy 2.(2n+1) chia hết cho 2n+1
=> 11 chia hết cho 2n+1
=> 2n+1 \(\in\)Ư( 11 ) ={ 1, -1, 11, -11}
+) 2n+1 = 1 => 2n= 0 => n =0
+) 2n+1 = -1 => 2n =-2 => n=-1
+) 2n+1 =11 => 2n = 10 => n=5
+) 2n+1 = -11 => 2n = -12 => n = -6
vậy n \(\in\){ 0,-1 , 5, -6}
4n+13 chia hết cho 2n+1 =>\(\frac{4n+13}{2n+1}\in Z\)
=> \(\frac{2n+1+2n+1+11}{2n+1}\in Z\)
=>\(2+\frac{11}{2n+1}\in Z\) =>\(\frac{11}{2n+1}\in Z\) => 2n+1 \(\in\) Ư(11)= { -11; -1; 1; 11}
=> 2n= -12; -2; 0; 10 => n= -6;-1;0;5