ĐKXĐ: \(x\in R\)
\(x^2+6x+8=2\left(x+2\right)\cdot\sqrt{x^2+5}\)
=>\(\left(x+2\right)\left(x+4\right)-2\left(x+2\right)\sqrt{x^2+5}=0\)
=>\(\left(x+2\right)\left[\left(x+4\right)-2\sqrt{x^2+5}\right]=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x+4-2\sqrt{x^2+5}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\left(nhận\right)\\x+4=\sqrt{4x^2+20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}4x^2+20=\left(x+4\right)^2\\x>=-4\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}4x^2+20-x^2-8x-16=0\\x>=-4\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}3x^2-8x+4=0\\x>=-4\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}\left(x-2\right)\left(3x-2\right)=0\\x>=-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x\in\left\{-2;2;\dfrac{2}{3}\right\}\)
