e)
x^2-6x+9\ne0=>(x-3)^2\ne 0=> x\ne3
D=R \ {3}
\frac{2x^2-3x}{x^2-6x+9}-\frac{x+2}{x-3}+1=\frac{2x^2-3x-(x+2)(x-3)+(x-3)^2}{(x-3)^2}=
\frac{2x^2-3x-(x^2-3x+2x-6)+x^2-6x+9}{(x-3)^2}=\frac{3x^2-9x+9-x^2+x+6}{(x-3)^2}=\frac{2x^2-8x+15}{(x-3)^2}
f)
x-1\ne0=>x\ne1 ; x^2-2x+1\ne0=>(x-1)^2\ne0=>x\ne1 ; x\ne0
D=R \ {0,1}
\frac{x+2}{x-1}-\frac{4}{x^2-2x+1}-\frac{x+1}{x}=\frac{x+2}{x-1}-\frac{4x}{(x-1)^2}-\frac{x+1}{x}=
\frac{x(x+2)(x-1)-4x-(x+1)(x-1)^2}{x(x-1)^2}=
\frac{(x^2+2x)(x-1)-4x-(x+1)(x^2-2x+1)}{x(x-1)^2}=
\frac{x^3-x^2+2x^2-2x-4x-(x^3-2x^2+x+x^2-2x+1)}{x(x-1)^2}=
\frac{x^3+2x^2-6x-(x^3-x^2-x+1)}{x(x-1)^2}=
\frac{x^3+2x^2-6x-x^3+x^2+x-1}{x(x-1)^2}=\frac{2x^2-5x-1}{x(x-1)^2}