Question

Thực hiện phép cộng :

\(\frac{x^2}{x^2-4}+\frac{1}{x+2}+\frac{2}{2-x}\)

 

Answer 1

ĐKXĐ\(\begin{cases}x^2-4\ne0\\x+2\ne0\\2-x\ne0\end{cases}\)<=>\(\begin{cases}x\ne-2\\x\ne2\end{cases}\)

Với \(x\ne-2,x\ne2\) ta có

        \(\frac{x^2}{x^2-4}+\frac{1}{x+2}+\frac{2}{2-x}\)=\(\frac{x^2}{\left(x+2\right)\left(x-2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\)

      =\(\frac{x^2}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

     =\(\frac{x^2+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\)=\(\frac{x^2-x+6}{\left(x+2\right)\left(x-2\right)}\)

     =\(\frac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-2\right)}\)=\(\frac{x-3}{x-2}\)