Question

Tính

\(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}+\dfrac{8}{x^2-4}\right):\dfrac{4}{x+2}\)

Answer 1

\(=\left(\dfrac{x+2}{2\left(x-2\right)}-\dfrac{x-2}{2\left(x+2\right)}+\dfrac{8}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4}{x+2}\)

\(=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2+8.2}{2\left(x-2\right)\left(x+2\right)}\)\(.\dfrac{x+2}{4}\)

=\(=\dfrac{\left(x+2+x-2\right)\left(x+2-x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{4}\)

=\(=\dfrac{2x.4+16}{2\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{4}\)

\(=\dfrac{8x+16}{2\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{4}\)

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

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

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

Chúc bạn học tốt!

Answer 2

Chúc bạn học tốt ^^

\(\left(\dfrac{x+2}{2\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{2\left(x+2\right)\left(x-2\right)}+\dfrac{16}{2\left(x^2-4\right)}\right):\dfrac{4}{x+2}=\dfrac{10}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{4}=\dfrac{5}{2\left(x-2\right)}\\ \)