Bài làm:
đk: \(x\ge3\)
Pt <=> \(\left(\sqrt{x-\sqrt{2x-5}-4}+\sqrt{x+\sqrt{2x-5}-4}\right)^2=\left(\sqrt{2}\right)^2\)
<=> \(x-\sqrt{2x-5}-4+x+\sqrt{2x-5}-4+2\sqrt{\left(x-4\right)^2-2x+5}=2\)
<=> \(2x-10=-2\sqrt{x^2-4x+4-2x+5}\)
<=> \(2x-10+2\sqrt{x^2-6x+9}=0\)
<=> \(2x-10+2\sqrt{\left(x-3\right)^2}=0\)
<=> \(2\left|x-3\right|=10-2x\)
<=> \(\left|x-3\right|=5-x\Leftrightarrow\orbr{\begin{cases}x-3=5-x\\x-3=x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=8\\0x=-2\left(∄x\right)\end{cases}\Rightarrow}x=4\)
