ĐKXĐ: \(x\ne0\)
Ta có: \(\frac{x-1}{x^2-x+1}+\frac{x+1}{x^2+x+1}=\frac{10}{x\left(x^4+x^2+1\right)}\)
\(=\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}+\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{10}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
Suy ra: \(x\left(x-1\right)\left(x^2+x+1\right)+x\left(x+1\right)\left(x^2-x+1\right)=10\)
\(\Leftrightarrow x\left(x^3-1\right)+x\left(x^3+1\right)-10=0\)
\(\Leftrightarrow x^4-x+x^4+x-10=0\)
\(\Leftrightarrow2x^4=10\)
\(\Leftrightarrow x^4=5\)
\(\Leftrightarrow x=\pm\sqrt[4]{5}\)
Vậy: \(x=\pm\sqrt[4]{5}\)
