a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)\(A=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{x^2-9}:\dfrac{2x+6-x-5}{x+3}\)\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)\(=\dfrac{-5\left(x+1\right)}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=-\dfrac{5}{x-3}\)b: \(x^2-x-2=0\)=>\(\left(x-2\right)\left(x+1\right)=0\)=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)Khi x=2 thì \(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)c: A=1/2=>\(-\dfrac{5}{x-3}=\dfrac{1}{2}\)=>x-3=-10=>x=-7(nhận)Câu trả lời: a: ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)\(A=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{x^2-9}:\dfrac{2x+6-x-5}{x+3}\)\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)\(=\dfrac{-5\left(x+1\right)}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=-\dfrac{5}{x-3}\)b: \(x^2-x-2=0\)=>\(\left(x-2\right)\left(x+1\right)=0\)=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)Khi x=2 thì \(A=\dfrac{-5}{2-3}=\dfrac{-5}{-1}=5\)c: A=1/2=>\(-\dfrac{5}{x-3}=\dfrac{1}{2}\)=>x-3=-10=>x=-7(nhận)