a, Ta có :
\(P=\dfrac{x+2}{x-3}\)
Để \(P=0\) thì :
\(x+2=0\Leftrightarrow x=-2\)
b, Để \(P< 0\Leftrightarrow\left\{{}\begin{matrix}x+2\\x-3\end{matrix}\right.\) cùng dấu
c, Để \(P>0\Leftrightarrow\left\{{}\begin{matrix}x+2\\x-3\end{matrix}\right.\) khác dấu
Điều kiện: x - 3 \(\ne\) 0
*Để P = 0 thì: x + 2 = 0
\(\Rightarrow\) x = -2
*Để P > 0 thì: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>3\\x< -2\end{matrix}\right.\)
*Để P < 0 thì: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\left(\text{loại}\right)\end{matrix}\right.\)
\(P=\dfrac{x+2}{x-3}\)
\(P=0\Rightarrow x+2=0\Rightarrow x=-2\)
\(P>0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2>0\Rightarrow x>-2\\x-3>0\Rightarrow x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x+2< 0\Rightarrow x< -2\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x>-2;x< 3\)
\(P< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\Rightarrow x< -2\\x-3>0\Rightarrow x>3\end{matrix}\right.\\\left\{{}\begin{matrix}x+2>0\Rightarrow x>-2\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow-2< x< 3\)
