Question

  cứu khổ cứu nạn dùm mình

Answer 1

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

\(D=R\backslash\left\{-1\right\}\)

\(TCĐ:x=-1\)

\(\)\(TCX:y=x\)

\(y'=\dfrac{\left(2x+1\right)\left(x+1\right)-\left(x^2+x-2\right)}{\left(x+1\right)^2}=\dfrac{x^2+2x+3}{\left(x+1\right)^2}>0,\forall x\in D\)

\(\RightarrowĐTHS\) đồng biến trên \(D\)