\(\dfrac{x-2}{x+3}=\dfrac{x+3-5}{x+3}=\dfrac{x+3}{x+3}-\dfrac{5}{x+3}=1-\dfrac{5}{x+3}.\text{Đ}k:x\ne-3\\ x-2⋮x+3\)
\(\Leftrightarrow\dfrac{5}{x+3}\)có giá trị nguyên
\(\Rightarrow x+3\in\text{Ư}\left(5\right)=\left\{\pm1;\pm5\right\} \)
+) \(x+3=1\Leftrightarrow x=1-3\Leftrightarrow x=-2\)
+)\(x+3=-1\Leftrightarrow x=\left(-1\right)-3\Leftrightarrow x=-4\)
+)\(x+3=5\Leftrightarrow x=5-3\Leftrightarrow x=2\)
+)\(x+3=-5\Leftrightarrow x=\left(-5\right)-3\Leftrightarrow x=-8\)
Ta có: \(\dfrac{x-2}{x+3}=\dfrac{x+3-5}{x+3}=1-\dfrac{5}{x+3}\)
Để \(\left(x-2\right)⋮\left(x+3\right)\) thì \(5⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{-1;1;-5;5\right\}\)
*) \(x+3=-5\)
\(x=-8\) (thỏa mãn)
*) \(x+3=-1\)
\(x=-4\) (thỏa mãn)
*) \(x+3=1\)
\(x=-2\) (thỏa mãn)
*) \(x+3=5\)
\(x=2\) (thỏa mãn)
Vậy \(x=-8;x=-4;x=-2;x=2\)
