\(a,=\dfrac{x^2+x-2x+2-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}\\ b,=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\\ c,=\dfrac{x^2\left(x+1\right)-x^3\left(x-1\right)-x^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x^3+x^2-x^4+x^3-x^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{-x^3\left(x-2\right)}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x^2\left(2-x\right)}{\left(x-1\right)\left(x+1\right)}\)
\(d,=\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}=\dfrac{4x}{2-x}\\ e,=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{4\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{4}{2x+1}\\ f,=\dfrac{9-x^2+x^2-9-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{2-x}{x+3}\)
