\(a,=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}=\dfrac{\left(x-1\right)^2}{x\left(x+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}=\dfrac{1}{x+1}\\ b,=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\cdot\dfrac{\left(1-3x\right)^2}{2x\left(3x+5\right)}\\ =\dfrac{x\left(3x+5\right)}{2x\left(3x+5\right)\left(3x+1\right)}=\dfrac{1}{2\left(3x+1\right)}\\ c,=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}=\dfrac{-3}{x-3}\\ d,=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
