\(C=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}\) \(=\frac{x^3-x^2-4x+4}{x^2-4}\)\(=\frac{\left(x+2\right)\left(x-2\right)\left(x-1\right)}{\left(x+2\right)\left(x-2\right)}=x-1\)
ĐKXĐ: \(x\ne\pm2\)
1) Để \(C=0\) \(\Leftrightarrow x-1=0\Leftrightarrow x=1\) (Thỏa mãn)
Vậy \(x=1\) thì \(C=0\)
2) Để C thuộc N*