Question

Cho \(\left(\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}+\dfrac{1}{x+2}\right)\left(\dfrac{2}{x}-1\right)\)
a) rút gọn
b) tìm x để A =1

Answer 1

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

\(=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\left(\dfrac{2-x}{x}\right)\)

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

\(=\left(-\dfrac{1}{x}+\dfrac{-2}{x+2}+\dfrac{2-x}{x\left(x+2\right)}\right)\)

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

b) Ta có ĐKXĐ của A là \(x\ne\pm2\)

Lại có \(A=-\dfrac{4}{x+2}=1\)

\(\Rightarrow-4=x+2\Rightarrow x=-6\)

Vậy x=-6 thì A=1