Question

giải phương trình :\(\sqrt{2x+3}=1+\sqrt{2}\)

Answer 1

ĐK: \(2x+3\ge0\Leftrightarrow x\ge-\frac{3}{2}\)

\(\sqrt{2x+3}=1+\sqrt{2}\\ \Leftrightarrow\left|2x+3\right|=1+2\sqrt{2}+2\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=3+2\sqrt{2}\\2x+3=-3-2\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+2\sqrt{2}+3}{2}=\sqrt{2}\left(TM\right)\\x=\frac{-3-2\sqrt{2}}{2}\left(KTM\right)\end{matrix}\right.\\ \Rightarrow S=\left\{\sqrt{2}\right\}\)