\(2x-1⋮x+3\)
\(\Leftrightarrow\) \(2\left(x+3\right)-4⋮x+3\)
\(\Rightarrow\) \(4⋮x+3\)
\(\Rightarrow\)\(x+3\inƯ\left(4\right)\)
\(\text{Ta có}\) \(Ư\left(4\right)=\left(-1;-2;-4;1;2;4\right)\)
\(\text{*Nếu x + 3 = -1}\)
\(\Rightarrow x=-4\)
\(\text{*Nếu x + 3 = -2}\)
\(\Rightarrow x=-5\)
\(\text{*Nếu x + 3 = -4
}\)
\(\Rightarrow x=-7\)
\(\text{*Nếu x + 3 = 1
}\)
\(\Rightarrow x=-2\)
\(\text{*Nếu x + 3 = 2}\)
\(\Rightarrow x=-1\)
\(\text{*Nếu x + 3 = 4}\)
\(\Rightarrow x=1\)
\(x\)\(\in\left\{-4;-5;-7;-2;-1;1\right\}\)
2x-1\(⋮\)x+3
2x+6-7\(⋮\)x+3
2(x+3)-7\(⋮\)x+3
Vì 2(x+3)\(⋮\)x+3
buộc 7\(⋮\)x+3=>x+3\(\in\)Ư(7)={1;7;-1;-7}
Với x+3=1=>x= -2
x+3=7=>x=4
x+3= -1=>x= -4
x+3= -7=>x= -10
Vậy x\(\in\){-2;4;-4;-10}