ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}-\frac{2\left(x^2+4x+4\right)^2}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-1\right)\left(x^2-x+1\right)}{\left(x^2+x+1\right)\left(x^2-1\right)\left(x^2-x+1\right)}-\frac{\left(x-1\right)\left(x^2-1\right)\left(x^2+x+1\right)}{\left(x^2-x+1\right)\left(x^2-1\right)\left(x^2+x+1\right)}-\frac{2x^2+8x+8}{\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2-1\right)\left(x^2+x+1\right)-\left(2x^2+8x+8\right)=0\)\(\Leftrightarrow x^5-x^3+x^2-1-\left(x^5-x^3-x^2+1\right)-2x^2-8x-8=0\)
\(\Leftrightarrow x^5-x^3+x^2-1-x^5+x^3+x^2-1-2x^2-8x-8=0\)
\(\Leftrightarrow-8x-10=0\)
\(\Leftrightarrow-\left(8x+10\right)=0\)
\(\Leftrightarrow8x+10=0\)
\(\Leftrightarrow8x=-10\)
\(\Leftrightarrow x=\frac{-10}{8}=-\frac{5}{4}\)
Vậy: \(x=-\frac{5}{4}\)
