ta có: \(\frac{x+3}{x+1}=\frac{x+1+2}{x+1}=\frac{x+1}{x+1}+\frac{2}{x+1}=1+\frac{2}{x+1}\)
Để x + 3/x+1 đạt giá trị nguyên
=> 2/x+1 có giá trị nguyên
=> 2 chia hết cho x + 1
=> x + 1 thuộc Ư(2)={1;-1;2;-2}
nếu x + 1 = 1 => x = 0 (TM)
x + 1 = -1 => x = -2 (TM)
x + 1 = 2 => x = 1 (TM)
x + 1 = - 2 => x = -3 (TM)
KL:...
Dể \(\frac{x+3}{x+1}\)là giá trị nguyên
=> \(x+3\)\(⋮\)\(x+1\)
=> \(x+1+2\)\(⋮\)\(x+1\)
=> Vì \(x+1\)\(⋮\)\(x+1\)nên \(2\)\(⋮\)\(x+1\)
=> \(x+1\)\(\in\)\(Ư\left(2\right)\)
=> \(x+1\)\(\in\)\(\left\{\pm1;\pm2\right\}\)
=> \(x\)\(\in\)\(\left\{0;-2;1;-3\right\}\)