Đáp án: $x=\dfrac32$
Giải thích các bước giải:
Ta có :
$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+x^2+1)}$
$\to\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+2x^2+1-x^2)}$
$\to\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x((x^2+1)^2-x^2)}$
$\to\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^2+1-x)(x^2+1+x)}$
$\to\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^2-x+1)(x^2+x+1)}$
$\to (x+1)(x^2-x+1)-(x-1)(x^2+x+1)=\dfrac3x$
$\to x^3+1-(x^3-1)=\dfrac3x$
$\to 2=\dfrac3x$
$\to x=\dfrac32$
