Question

cho hàm số \(y=x+\sqrt{x+x^2}\). Tìm tập nghiệm bpt \(x.y'>y\)

Answer 1

\(y'=1+\dfrac{1}{2\sqrt{x^2+x}}\left(2x+1\right)\)

\(\Rightarrow xy'=x+\dfrac{2x^2+x}{2\sqrt{x^2+x}}>x+\sqrt{x+x^2}\)

\(\Leftrightarrow\dfrac{2x^2+x}{2\sqrt{x^2+x}}>\sqrt{x^2+x}\Leftrightarrow2x^2+x>2\left(x^2+x\right)\Leftrightarrow2x^2+x>2x^2+2x\Leftrightarrow x< 0\)

\(\Rightarrow S=\left(-\infty;0\right)\)