Question

Cho x>1, so sánh \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)với\(\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}\)

Answer 1

Giả sử \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}>\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1\right]>0\left(1\right)\)

Với \(x>1\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>1>0\\\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1>0\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) luôn luôn đúng \(\forall x>1\)

Vậy \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}\)