Question

Cho A = ( \(\frac{4x}{x^2-4}\) + \(\frac{2x-4}{x+2}\) ) . \(\frac{x+2}{2x}\) + \(\frac{2}{2-x}\) ( x khác 0 , x khác -2 , x khác 2 )

a) Rút gọn A

b) Tính A khi x = 4

c) Tìm giá trị nguyên của x để A nhận nguyên

Answer 1

a: Ta có: \(A=\left(\dfrac{4x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{x+2}\right)\cdot\dfrac{x+2}{2x}-\dfrac{2}{x-2}\)

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

\(=\dfrac{4x+2x^2-8x+8}{x-2}\cdot\dfrac{1}{2x}-\dfrac{2}{x-2}\)

\(=\dfrac{2x^2-12x+8}{2x\left(x-2\right)}-\dfrac{2}{x-2}\)

\(=\dfrac{2x^2-12x+8-4x}{2x\left(x-2\right)}=\dfrac{2x^2-16x+8}{2x\left(x-2\right)}\)

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

b: Thay x=4 vào A, ta được:

\(A=\dfrac{4^2-8\cdot4+4}{4\cdot\left(4-2\right)}=\dfrac{-12}{4\cdot2}=\dfrac{-12}{8}=-\dfrac{3}{2}\)