a) \(A=\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\)
\(A=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\)
b) \(A:B=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}:\left(2-\dfrac{x+5}{x+3}\right)\)
\(A:B=\dfrac{-5\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-x-5}{x+3}\)
\(A:B=\dfrac{-5\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{-2}\)
\(A:B=\dfrac{5\left(x+1\right)}{2\left(x-3\right)}=\dfrac{1}{2}\)
\(\Rightarrow10x+10=2x-6\)
\(\Rightarrow10x-2x=-6-10\)
\(\Rightarrow8x=-16\)
\(\Rightarrow x=-2\)
c) \(x^2-x-2=0\Rightarrow x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+\left(x-2\right)\Rightarrow\left(x+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) Khi x = -1 ta có:
\(A=\dfrac{-5\cdot-1-5}{\left(-1-3\right)\left(-1+3\right)}=\dfrac{5-5}{-4\cdot2}=0\)
Khi x = 2 ta có:
\(A=\dfrac{-5\cdot2-5}{\left(2-3\right)\left(2+3\right)}=\dfrac{-10-5}{-1\cdot5}=\dfrac{-15}{-5}=3\)
