Giải phương trình: x 4x ----- + ----------- = 0 y x ² - 4 help vs ạ ><

$\dfrac{x}{x+2}+\dfrac{4x}{x^2-4}=0$

$ĐKXĐ:$

$⇒$ \(\left[ \begin{array}{l}x+2≠0\\x^2-4≠0\end{array} \right.\) 

$⇒$ \(\left[ \begin{array}{l}x≠-2\\(x-2)(x+2)≠0\end{array} \right.\) 

$⇒$ \(\left[ \begin{array}{l}x≠-2\\x≠±2\end{array} \right.\) 

$⇒ x≠±2$

$⇒ \dfrac{x}{x+2}+\dfrac{4x}{(x-2)(x+2)}=0$

$⇒ \dfrac{x(x-2)}{(x-2)(x+2)}+\dfrac{4x}{(x-2)(x+2)}=0$

$⇒ \dfrac{x^2-2x}{(x-2)(x+2)}+\dfrac{4x}{(x-2)(x+2)}=0$

$⇒ \dfrac{x^2-2x+4x}{(x-2)(x+2)}=0$

$⇒ \dfrac{x^2+2x}{(x-2)(x+2)}=0$

$⇒ \dfrac{x(x+2)}{(x-2)(x+2)}=0$

$⇒ \dfrac{x}{x-2}=0$

$⇒ x=0$

Vậy $x=0(t/m ĐKXĐ)$

$#thanhmaii2008$