\(\sqrt{2x+3}=1+\sqrt{2}\left(ĐK:x\ge-\dfrac{3}{2}\right)\)
<=> \(\left|2x+3\right|=\left(1+\sqrt{2}\right)^2\)
<=> \(\left|2x+3\right|=3+2\sqrt{2}\)
<=> \(\left[{}\begin{matrix}2x+3=3+2\sqrt{2}\\-2x-3=3+2\sqrt{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=3-\sqrt{2}\end{matrix}\right.\)