Question

Thực hiện phép tính sau :

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

Answer 1

ĐKx\(\ne\)2,x\(\ne\)0

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

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

=\(\frac{x+2}{x-2}\)

Answer 2

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

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

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

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

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

\(=\frac{x+2}{x-2}\)