${\dfrac{x+3}{x-3}+\dfrac{4x^2}{9-x^2}=\dfrac{x-3}{x+3}}$ $\text{ ĐKXĐ}$ ${ x \neq 3; x \neq -3}$
⇔ ${\dfrac{(x+3).(x+3)}{(x-3).(x+3)}-\dfrac{4x^2}{(x-3).(x+3)}=\dfrac{(x-3).(x-3)}{(x+3).(x-3)}}$
⇔ ${\dfrac{x^2+6x+9}{(x-3).(x+3)}-\dfrac{4x^2}{(x-3).(x+3)}=\dfrac{x^2-6x+9}{(x-3).(x+3)}}$
⇔ ${x^2+6x+9-4x^2=x^2-6x+9}$
⇔ ${ x^2+6x+9-4x^2-x^2+6x-9=0}$
⇔ ${ -4x^2+12x=0}$
⇔ ${ -4x.(x-3)=0}$
⇔ \(\left[ \begin{array}{l}-4x=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\text{(thỏa mãn ĐKXĐ)}\\x=3\text{(không thỏa mãn ĐKXĐ)}\end{array} \right.\)
$\text{Vậy phương trình đã cho có nghiệm x =0}$
