Đk: x ≠ -3; x≠1; x≠-2
A = \(\frac{\left(x+3\right)\left(x+2\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)}=\frac{1}{x-1}\)
1/ \(A>\frac{3}{x-1}\Leftrightarrow\frac{1}{x-1}>\frac{3}{x-1}\Leftrightarrow\frac{-2}{x-1}>0\Leftrightarrow x-1< 0\Leftrightarrow x< 0\)
Kết howpj đk =< x<0 và x ≠ -3; x≠-2
2/A(x+2) = \(\frac{x+2}{x-1}=\frac{x-1+3}{x-1}=1+\frac{3}{x-1}\)
để biểu thức là số tự nhiên
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\inƯ\left(3\right)\\\frac{3}{x-1}\ge-1\end{matrix}\right.\Leftrightarrow\)\(\left\{{}\begin{matrix}x-1=\left\{-3;-1;1;3\right\}\\\frac{3}{x-1}\ge-1\end{matrix}\right.\)
\(\Leftrightarrow x=\left\{-2;2;4\right\}\)
