Question

rút gọn bài 8 hộ mình vs ạ

Answer 1

Bài 8:

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

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

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

b: Thay x=3 vào E, ta được:

\(E=\dfrac{4\cdot3}{\left(3+1\right)^2}=\dfrac{12}{4^2}=\dfrac{3}{4}\)

Thay x=-3 vào E, ta được:

\(E=\dfrac{4\cdot\left(-3\right)}{\left(-3+1\right)^2}=\dfrac{-12}{4}=-3\)