ĐKXĐ: \(x\ge-8\)
\(\left(2x+1\right)^2-2\left(2x+1\right)\sqrt{x+8}+\left(x+8\right)-x^2+2x-1=0\)
\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x+2-\sqrt{x+8}\right)\left(3x-\sqrt{x+8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+8}=x+2\\\sqrt{x+8}=3x\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x\ge-2\\x+8=\left(x+2\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+3x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\left(l\right)\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x\ge0\\x+8=9x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge0\\9x^2-x-8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8}{9}\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=1\)
