Question

\(\dfrac{x+2}{x-2}\)-\(\dfrac{x-2}{x+2}\)=\(\dfrac{16}{x^2-4}\)

Answer 1

đk x khác -2;2 

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{16}{x^2-4}\)

\(\Rightarrow\left(x+2-x+2\right)\left(x+2+x-2\right)=16\Leftrightarrow2x=4\Leftrightarrow x=2\)(ktm) 

Vậy pt vô nghiệm 

Answer 2

ĐKXĐ: $x\ne\pm2$

$\frac{x+2}{x-2}-\frac{x-2}{x+2}=\frac{16}{x^2-4}$

$\Leftrightarrow \frac{(x+2)^2}{(x-2)(x+2)}-\frac{(x-2)^2}{(x-2)(x+2)}=\frac{16}{(x-2)(x+2)}$

$\Rightarrow (x+2)^2-(x-2)^2=16$

$\Leftrightarrow x^2+4x+4-x^2+4x-4=16$

$\Leftrightarrow 8x=16$

$\Leftrightarrow x=2$ (tm ĐKXĐ)

#$\mathtt{Toru}$