\(a,=\dfrac{x^2+2xy-y^2}{xy\left(x-y\right)}=\dfrac{\left(x-y\right)^2}{xy\left(x-y\right)}=\dfrac{x-y}{xy}\\ b,=\dfrac{x-1-x-1+2x^2}{\left(x+1\right)\left(x-1\right)}=\dfrac{2\left(x^2-1\right)}{x^2-1}=2\\ c,=\dfrac{x^2+xy+x^2-xy+2y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x^2+2y^2}{\left(x-y\right)\left(x+y\right)}\\ d,=\dfrac{x^2+2x+1+x^2-2x+1-2x^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x^2-1}\\ e,=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{x^2+x+1}\)
\(f,=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
