Giải thích các bước giải:
$\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}$
$=\dfrac{\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right)\left(1-6x+9x^2\right)}{6x^2+10x}$
$=\dfrac{\left(\dfrac{3x(3x+1)+2x(1-3x)}{(1-3x)(3x+1)}\right)\left(1-3x\right)^2}{6x^2+10x}$
$=\dfrac{\left(\dfrac{3x^2+5x}{3x+1}\right).\left(1-3x\right)}{2x(3x+5)}$
$=\dfrac{\left(\dfrac{x(3x+5)}{3x+1}\right).\left(1-3x\right)}{2x(3x+5)}$
$=\dfrac{1-3x}{2(3x+1)}$
