Đáp án: $x=\dfrac32$
Giải thích các bước giải:
Ta có:
$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+2x^2+1-x^2)}$
$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x((x^2+1)^2-x^2)}$
$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^2+1-x)(x^2+1+x)}$
$\to \dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^2-x+1)(x^2+x+1)}$
$\to x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)=3$
$\to x(x^3+1)-x(x^3-1)=3$
$\to (x^4+x)-(x^4-x)=3$
$\to 2x=3$
$\to x=\dfrac32$
