Đáp án:
$86) \dfrac{4-20x^2}{1-4x^2} - \dfrac{7}{2x+1} = \dfrac{3}{2x-1}$
$\text{ĐKXĐ : x $\neq$ ± $\dfrac{1}{2}$ }$
$⇔ \dfrac{20x^2-4}{4x^2-1} - \dfrac{7}{2x+1} = \dfrac{3}{2x-1}$
$⇔ \dfrac{20x^2-4}{(2x-1)(2x+1)} - \dfrac{7(2x-1)}{(2x+1)(2x-1)} = \dfrac{3(2x+1)}{(2x-1)(2x+1)}$
$⇔ 20x^2-4 - 7(2x-1) = 3(2x+1)$
$⇔ 20x^2 -4 - 14x +7 = 6x +3$
$⇔ 20x^2 -14x -6x -4+7-3=0$
$⇔ 20x^2 -20x=0$
$⇔ 20x(x-1)=0$
⇔\(\left[ \begin{array}{l}20x=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0(TM)\\x=1(TM)\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S={0 ; 1}}$
