\(a,=\dfrac{x^2-3-x-1+2-x^2}{x-2}=\dfrac{-x-2}{x-2}=\dfrac{x+2}{2-x}\\ b,=\dfrac{4-x^2-2x+x^2+x}{x-5}=\dfrac{4-x}{x-5}\\ c,=\dfrac{x-3+3x+9-5x}{\left(x-3\right)\left(x+3\right)}=\dfrac{6-x}{\left(x-3\right)\left(x+3\right)}\\ d,=\dfrac{x^2+xy+x^2-xy-4}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x^2-4}{x^2-y^2}\\ e,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ f,=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
a: \(=\dfrac{x^2-3-x-1+2-x^2}{x-2}=\dfrac{-x+2}{x-2}=-1\)
