điều kiện \(\left\{{}\begin{matrix}x>2\\hoặc\\x< -2\end{matrix}\right.\)
\(\sqrt{x^2-4}-\sqrt{x-2}>0\Leftrightarrow\sqrt{x^2-4}>\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}>\sqrt{x-2}\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}>\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x+2}>\dfrac{\sqrt{x-2}}{\sqrt{x-2}}\Leftrightarrow\sqrt{x+2}>1\Leftrightarrow x+2>1\)
\(\Leftrightarrow x>1-2\Leftrightarrow x>-1\) chỉ thỏa mãng khi \(x>2\) (vì đk : \(\left\{{}\begin{matrix}x>2\\hoặc\\x< -2\end{matrix}\right.\))
vậy \(x>2\) thì \(\sqrt{x^2-4}-\sqrt{x-2}>0\)
\(\sqrt{x^2-4}-\sqrt{x-2}>0\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\sqrt{x^2-4}>\sqrt{x-2}\)
\(\Leftrightarrow x^2-4>x-2\)
\(\Leftrightarrow x^2-x-2>0\)
\(\Leftrightarrow x^2+x-2x-2>0\)
\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)>0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\left(ktm\right)\\x< -1\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x>2\left(tm\right)\)
Vậy x>2 thỏa bpt trên.
