Giải thích các bước giải:
c.Ta có :
$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{(x+1)(x^2-x+1)-(x-1)(x^2+x+1)}{(x^2+x+1)(x^2-x+1)}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{(x^3+1)-(x^3-1)}{(x^2+1)^2-x^2}=\dfrac{3}{x(x^4+x^2+1)}$
$\to \dfrac{2}{(x^2+1)^2-x^2}=\dfrac{3}{x(x^4+x^2+1)}$
$\to x=\dfrac32$
d.Ta có :
$\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}$
$\dfrac{12}{1-9x^2}=\dfrac{(1-3x)^2-(1+3x)^2}{(1+3x)(1-3x)}$
$\dfrac{12}{1-9x^2}=\dfrac{-12x}{1-9x^2}$
$\to x=-1$
