$\sqrt{2x+8-6\sqrt{2x-1} } + \sqrt{2x+2\sqrt{2x-1} } = 4$
$\sqrt{(2x-1)-6\sqrt{2x-1}+9 } + \sqrt{(2x-1)+2\sqrt{2x-1}+1 } = 4$
$\sqrt{(\sqrt{2x-1}-3)^2 } + \sqrt{(\sqrt{2x-1}+1)^2 } = 4$
$|\sqrt{2x-1}-3| + \sqrt{2x-1}+1 = 4$
กรณี $0\leqslant \sqrt{2x-1}<3 \Rightarrow \frac{1}{2}\leqslant x < 5 $
$3-\sqrt{2x-1} + \sqrt{2x-1}+1 = 4$
$4 = 4$
$\therefore $ กรณีนี้เป็นจริงทุก $x \in [\frac{1}{2}, 5)$
กรณี $3\leqslant \sqrt{2x-1} \Rightarrow 5\leqslant x $
$\sqrt{2x-1}-3 + \sqrt{2x-1}+1 = 4$
$2\sqrt{2x-1} = 6$
$x = 5$
$\therefore x \in [\frac{1}{2}, 5]$
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