$\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1} \left ( x \neq ± \dfrac{1}{4} \right )$
$⇔\dfrac{-3(4x+1)}{(4x-1)(4x+1)}=\dfrac{2(4x-1)}{(4x+1)(4x-1)}-\dfrac{8+6x}{(4x+1)(4x-1)}$
$⇒-3(4x+1)=2(4x-1)-(8+6x)$
$⇔-12x-3=8x-2-8-6x$
$⇔-12x-3=2x-10$
$⇔2x+12x=-3+10$
$⇔14x=7$
$⇔x=\dfrac{1}{2}$ ( Thỏa mãn )
Vậy phương trình có nghiệm $x=\dfrac{1}{2}$
