Question

Thực hiện phép tính:

B=(2/x-2 + 2x/ x+2) . x^2 +4x +4

                                         8

(x khác 0, x khác + hoặc  3)

 

 

Answer 1

Với \(x\ne0;x\ne\pm3\)

\(B=\left(\frac{2}{x-2}+\frac{2x}{x+2}\right).\frac{\left(x+2\right)^2}{8}\)

\(=\left(\frac{2\left(x+2\right)+2x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right).\frac{\left(x+2\right)^2}{8}\)

\(=\left(\frac{2x+4+2x^2-4x}{\left(x-2\right)}\right).\frac{x+2}{8}=\frac{2x^2-2x+4}{x-2}.\frac{x+2}{8}\)

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