x+11 chia hết cho x+1
=>(x+1)+10 chia hết cho x+1
Vì x+1 chia hết cho x+1
=>10 chia hết cho x+1
=>x+1 thuộc(1;2;5;10)
+,Nếu x+1=1=>x=0(TM)
+,Nếu x+1=2=>x=1(TM)
+,Nếu x+1=5=>x=4(TM)
+,Nếu x+1=10=>x=9(TM)
Vậy x={0;1;4;9}
\(\frac{x+11}{x+1}=\frac{x+1+10}{x+1}=\frac{x+1}{x+1}+\frac{10}{x+1}=1+\frac{10}{x+1}\)
Để (x + 11) \(⋮\) (x + 1) thì 10 \(⋮\) (x + 1)
\(\Rightarrow\) x + 1 = 1; x + 1 = -1; x + 1 = 2; x + 1 = -2; x + 1 = 5; x + 1 = -5; x + 1 = 10; x + 1 = -10
*) x + 1 = 1
x = 1 - 1
x = 0
*) x + 1 = -1
x = -1 - 1
x = -2
*) x + 1 = 2
x = 2 - 1
x = 1
*) x + 1 = -2
x = -2 - 1
x = -3
*) x + 1 = 5
x = 5 - 1
x = 4
*) x + 1 = -5
x = -5 - 1
x = -6
*) x + 1 = 10
x = 10 - 1
x = 9
*) x + 1 = -10
x = -10 - 1
x = -11
Vậy x = -11; x = 9; x = -6; x = 4; x = -3; x = 1; x = -2; x = 0 thì (x + 11) \(⋮\) (x + 1)