a) \(ĐKXĐ:\hept{\begin{cases}x\ne-2\\x\ne3\\x\ne2\end{cases}}\)
\(A=\left(1-\frac{4}{x+2}\right):\left(1+\frac{1}{x-3}\right)\)
\(\Leftrightarrow A=\frac{x-2}{x+2}:\frac{x-2}{x-3}\)
\(\Leftrightarrow A=\frac{x-3}{x+2}\)
b) Để A nguyên
\(\Leftrightarrow x-3⋮x+2\)
\(\Leftrightarrow x+2-5⋮x+2\)
\(\Leftrightarrow5⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1;-7;3\right\}\)
Vậy để A nguyên \(\Leftrightarrow x\in\left\{-3;1;-7;3\right\}\)
c) Để A > 0
\(\Leftrightarrow\frac{x-3}{x+2}>0\)
\(\Leftrightarrow1-\frac{5}{x+2}>0\)
\(\Leftrightarrow\frac{5}{x+2}< 0\)
\(\Leftrightarrow x+2< 0\)(vì 5 > 0)
\(\Leftrightarrow x< -2\)
Vậy để A > 0 \(\Leftrightarrow x< -2\)
