1: ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
2: \(M=\left(\dfrac{x}{x-3}-\dfrac{x+3}{3x^2-6x-9}+\dfrac{1}{3x+3}\right)\cdot\dfrac{x^2-2x-3}{x^2+x+2}\)
\(=\left(\dfrac{x}{x-3}-\dfrac{x+3}{3\left(x-3\right)\left(x+1\right)}+\dfrac{1}{3\left(x+1\right)}\right)\cdot\dfrac{\left(x-3\right)\left(x+1\right)}{x^2+x+2}\)
\(=\dfrac{3x\left(x+1\right)-x-3+x-3}{3\left(x-3\right)\left(x+1\right)}\cdot\dfrac{\left(x-3\right)\left(x+1\right)}{x^2+x+2}\)
\(=\dfrac{3x^2+3x-6}{3}\cdot\dfrac{1}{x^2+x+2}=\dfrac{x^2+x-2}{x^2+x+2}\)
3: Vì x^2+x-2<x^2+x+2
nên M<1