\(\frac{x+5}{x-2}\)có giá trị nguyên
\(\Leftrightarrow x+5⋮x-2\)
\(\Rightarrow\left(x-2\right)+7⋮x-2\)
có \(x-2⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2\inƯ\left(7\right)\)
\(x\in Z\Rightarrow x-2\in Z\)
\(\Rightarrow x-2\in\left\{-1;-7;1;7\right\}\)
\(\Rightarrow x\in\left\{1;-5;3;9\right\}\)
\(A=\frac{x+5}{x-2}\)
\(A=\frac{x-2+7}{x-2}\)
\(A=1+\frac{7}{x-2}\)
để \(A\in Z\)thì \(\frac{7}{x-2}\in Z\)
\(\Leftrightarrow x-2\inƯ\left(7\right)\)
\(\Leftrightarrow x-2\in\left\{\pm1;\pm7\right\}\)
đến đây tự làm