\(\dfrac{x^2}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x^2}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x^2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2\left(x-2\right)+4\left(x+2\right)=4\)
\(\Leftrightarrow x^3-2x^2+4x+8=4\)
\(\Leftrightarrow x^3-2x^2+4x+8-4=0\)
\(\Leftrightarrow x^3-2x^2+4x+4=0\)
PT vô nghiệm vì không thể tìm được x
Vậy : ....