\(\dfrac{1}{x^2-4x-5}=\dfrac{1}{\left(x^2+x\right)-\left(5x+5\right)}=\dfrac{1}{x\left(x+1\right)-5\left(x+1\right)}=\dfrac{1}{\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{2}{x^2-2x-x}=\dfrac{2}{x^2-3x}=\dfrac{2}{x\left(x-3\right)}\)
MTC \(x\left(x-3\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{1}{x^2-4x-5}=\dfrac{1}{\left(x^2+x\right)-\left(5x+5\right)}=\dfrac{1}{x\left(x+1\right)-5\left(x+1\right)}\\ =\dfrac{1}{\left(x+1\right)\left(x-5\right)}=\dfrac{x\left(x-3\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{x^2-3x}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{2}{x^2-2x-x}=\dfrac{2}{x^2-3x}=\dfrac{2}{x\left(x-3\right)}=\dfrac{2\left(x+1\right)\left(x-5\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{\left(2x+2\right)\left(x-5\right)}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\\ =\dfrac{2x^2-10x+2x-10}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}=\dfrac{2x^2-8x-10}{x\left(x-3\right)\left(x+1\right)\left(x-5\right)}\)