Giải giúp vs ạ

a) Ta có: $\frac{1}{x^2-x+1}:\frac{x+1}{x-1}=\frac{1}{x^2-x+1}.\frac{x-1}{x+1}=\frac{x-1}{(x^2-x+1)(x+1)}$ b) Ta có: $\frac{x+y}{2x-y}:\frac{1}{x-y}=\frac{x+y}{2x-y}.(x-y)=\frac{(x+y)(x-y)}{2x-y}=\frac{x^2-y^2}{2x-y}$ c) Ta có: $\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y}:(x^2+y^2)=\frac{x^2+y^2}{x^3}:(x^2+y^2)=\frac{x^2+y^2}{x^3}.\frac{1}{x^2+y^2}=\frac{1}{x^3}$ d) Ta có: $\frac{x^3+8}{x^2-2x+1}:\frac{x}{-1}=\frac{x^3+8}{x^2-2x+1}.\frac{-1}{x}=\frac{(x+2)(x^2-2x+4)}{(x-1)^2}.\frac{-1}{x}=\frac{-(x+2)(x^2-2x+4)}{x(x-1)^2}$