\(ĐKXĐ:x\ge-\frac{3}{2}\)
Ta có : \(2\sqrt{x+\sqrt{2x+3}+2}=\sqrt{2}\left(x+1\right)\)
\(\Leftrightarrow2.\sqrt{2}.\sqrt{x+\sqrt{2x+3}+2}=\sqrt{2}.\sqrt{2}.\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{2x+2\sqrt{2x+3}+4}=2\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{\left(2x+3\right)+2\sqrt{2x+3}+1}=2.\left(x+1\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x+3}+1\right)^2}=x+1\)
\(\Leftrightarrow\sqrt{2x+3}+1=x+1\)
\(\Leftrightarrow\sqrt{2x+3}=x\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\2x+3=x^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2-2x-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-3\right)\left(x+1\right)=0\end{cases}}\) \(\Leftrightarrow x=3\)( Thỏa mãn )
Vậy pt có nghiệm duy nhất \(x=3\)
