Question

[Umi Otaku]

ta có D =\(\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\left(\frac{2x+1}{x^2+x+1}\right)\)

         ( đkxđ: x khác 1 và -1, x khác -1/2)

   =\(\left(\frac{1}{x-1}+\frac{x}{ \left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\left(\frac{2x+1}{x^2+x+1}\right)\)

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

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

Answer 1

[ngonhuminh]

Tiếp

\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(\frac{x^2+x+1}{2x+1}\right)=\left(\frac{x^2+x+1}{x^2-1}\right)=1+\frac{x+2}{x^2-1}\)