ĐK: \(x ≥ \dfrac{1}{5}\)
\( \sqrt{5x-1}-\sqrt{x-1} > \sqrt{2x-4} \\ \Leftrightarrow \sqrt{5x-1} > \sqrt{2x-4}+\sqrt{x-1} \\ \Leftrightarrow 5x-1 > 2x-4+x-1-2\sqrt{(2x-4)(x-1)} \\ \Leftrightarrow 2x +6 > 2\sqrt{(2x-4)(x-1)} \\ \Leftrightarrow x+3 > \sqrt{(2x-4)(x-1)} \\ \Leftrightarrow (x+3)^2 > (2x-4)(x-1) \\ \Leftrightarrow x^2+6x+9 > 2x^2 -6x+4 \\ \Leftrightarrow x^2 -12x -5 < 0 \\ \Leftrightarrow 6-\sqrt{41} < x < 6+\sqrt{41}\)
Vậy \(\dfrac{1}{5} ≤ x < 6+\sqrt{41}\)
ĐK; \(x\ge2\)
\(\sqrt{5x-1}-\sqrt{x-1}>\sqrt{2x-4}\)
\(\Leftrightarrow\sqrt{5x-1}>\sqrt{2x-4}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1>2x-4+x-1+2\sqrt{\left(x-1\right)\left(2x-4\right)}\)
\(\Leftrightarrow x+2>\sqrt{\left(x-1\right)\left(2x-4\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2>0\\\left(x+2\right)^2>\left(x-1\right)\left(2x-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x^2-10x< 0\end{matrix}\right.\)
\(\Leftrightarrow0< x< 10\)
Vậy \(2\le x< 10\)
