Question

Cho biểu thức:

A=\(\left(\dfrac{x}{x^2-4}+\dfrac{1}{x+2}-\dfrac{2}{x-2}\right)\div\left(1-\dfrac{x}{x+2}\right)\)

(x\(\ne+2,-2\))

a)Rút gọn A

b)Tính giá trị của A khi x= -4

c)Tìm x ∈ z để A∈z

Answer 1

a) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{1}{x+2}-\dfrac{2}{x-2}\right):\left(1-\dfrac{x}{x+2}\right)\)

\(\Leftrightarrow\left(\dfrac{x}{x^2-4}+\dfrac{1\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x+2}{x+2}-\dfrac{x}{x+2}\right)\)\(\Leftrightarrow\)\(\dfrac{x+x-2-2x-4}{x^2-4}:\left(\dfrac{2}{x+2}\right)\)

\(\Leftrightarrow\dfrac{-6}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{2}\Leftrightarrow\dfrac{-3}{x-2}\)(kết quả cần tìm)

b) Khi x= -4

\(\Leftrightarrow\dfrac{-3}{4-2}=-\dfrac{3}{2}\)

Answer 2

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

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}=\dfrac{-3}{x-2}\)

b: Khi x=-4 thì \(A=\dfrac{-3}{-4-2}=\dfrac{-3}{-6}=\dfrac{1}{2}\)

c: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)