Giải thích các bước giải:
ĐKXĐ : $x\ne\pm2$
$\dfrac{-18x^2+5x+9}{x^2-4}=0$
$\to -18x^2+5x+9=0$
$\to -18\left(x-\dfrac{5}{36}\right)^2+\dfrac{673}{72}=0$
$\to 18\left(x-\dfrac{5}{36}\right)^2=\dfrac{673}{72}$
$\to \left(x-\dfrac{5}{36}\right)^2=\dfrac{673}{1296}$
$\to x=\dfrac{-\sqrt{673}+5}{36}\quad \mathrm{and}\quad \:x=\dfrac{\sqrt{673}+5}{36}$
