Question

Giải phương trình:

\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)

Answer 1

ĐK \(x\ne0\)

Ta có \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\cdot\left(x^4+x^2+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^4+x^2+1\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)

\(\Rightarrow\left(x^2+x\right)\left(x^2-x+1\right)-\left(x^2-x\right)\left(x^2+x+1\right)=3\)

\(\Leftrightarrow x^4-x^3+x^2+x^3-x^2+x-x^4-x^3-x^2+x^3+x^2+x=3\)

\(\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\left(tm\right)\)

Vậy \(x=\frac{3}{2}\)