Question

\(\frac{2x+1}{x+\sqrt{x+1}}\le1\)

Answer 1

ĐKXĐ: ...

Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x=t^2-1\)

\(\Rightarrow\frac{2\left(t^2-1\right)+1}{t^2+t-1}\le1\)

\(\Leftrightarrow\frac{2t^2-1}{t^2+t-1}-1\le0\)

\(\Leftrightarrow\frac{t^2-t}{t^2+t-1}\le0\Leftrightarrow\frac{t\left(t-1\right)}{t^2+t-1}\le0\Rightarrow\left[{}\begin{matrix}t=1\\\frac{\sqrt{5}-1}{2}\le t\le1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\frac{\sqrt{5}-1}{2}\le\sqrt{x+1}\le1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\\frac{1-\sqrt{5}}{2}\le x\le0\end{matrix}\right.\)