ĐKXĐ: \(x\ge\dfrac{1+\sqrt{17}}{8}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+\sqrt{x+1}+1}=a>0\\\sqrt{2x-\sqrt{x+1}}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2-b^2=2x+\sqrt{x+1}+1-2x+\sqrt{x+1}=2\sqrt{x+1}+1\)
Phương trình đã cho trở thành:
\(a+b=a^2-b^2\Leftrightarrow a+b=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b-1\right)=0\Leftrightarrow a=b+1\) (do \(a+b>0\))
\(\Leftrightarrow\sqrt{2x+\sqrt{x+1}+1}=\sqrt{2x-\sqrt{x+1}}+1\)
\(\Leftrightarrow2x+\sqrt{x+1}+1=2x-\sqrt{x+1}+1+2\sqrt{2x-\sqrt{x+1}}\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt{2x-\sqrt{x+1}}\)
\(\Leftrightarrow x+1=2x-\sqrt{x+1}\Leftrightarrow\sqrt{x+1}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x+1=\left(x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-3x=0\end{matrix}\right.\) \(\Rightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)