\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=-\dfrac{16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\dfrac{16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x^2+2x+2x+4-\left(x^2-2x-2x+4\right)+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+2x+2x-4+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8x+16}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{8}{x-2}=0\)
\(\Leftrightarrow8=0\) (vô lý)
S=\(\left\{\varnothing\right\}\)
Giải:
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-16}{x^2-4}\)
\(\Rightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-\left(x^2-4x+4\right)=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4=-16\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=-2\) (không thỏa mãn)
Vậy ...
