Question

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

Answer 1

\(=\dfrac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)+8}{x^2-4}\left(đk:x\ne\pm2\right)\)

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

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

Answer 2

xem lại đề bài 

Answer 3

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}+\dfrac{8}{x^2-4}=0\left(x\ne\pm2\right)\) (do ko có kết quả nên mik nghĩ là 0)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{x^2-4}-\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{8}{x^4-4}=0\)

\(\Rightarrow x^2-2x-x+2-x^2-2x+8=0\)

\(\Leftrightarrow-5x=-10\)

\(\Leftrightarrow x=2\left(loại\right)\)

Vậy pt có \(S=\varnothing\)