Question

\(\dfrac{1}{x^2-2x+1}\) - (\(\dfrac{x}{x^2-1}\) - \(\dfrac{1}{x^3-1}\)): \(\dfrac{x^2-2x+1}{x+x^3}\)

Answer 1

đk x khác 0 ; 1 

\(=\dfrac{1}{\left(x-1\right)^2}-\left[\dfrac{x\left(x^2+x+1\right)-x-1}{\left(x^2-1\right)\left(x^2+x+1\right)}\right]:\dfrac{\left(x-1\right)^2}{x\left(1+x^2\right)}\)

\(=\dfrac{1}{\left(x-1\right)^2}-\dfrac{x^3+x^2-1}{\left(x^2-1\right)\left(x^2+x+1\right)}:\dfrac{\left(x-1\right)^2}{x\left(x^2+1\right)}\)

\(=\dfrac{1}{\left(x-1\right)^2}-\dfrac{\left(x^3+x^2-1\right)x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^3-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)^2\left(x^2+x+1\right)-x\left(x^3+x^2-1\right)\left(x^2+1\right)}{\left(x-1\right)^2\left(x+1\right)^2\left(x^2+x+1\right)}\)