\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(\Leftrightarrow A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow A=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
Ta có : \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Khi x = 3, thì :
\(\dfrac{3-4}{3-2}=-1\)
Khi x = -3, thì :
\(\dfrac{-3-4}{-3-2}=\dfrac{-7}{-5}=\dfrac{7}{5}\)
Vậy .............
