ĐK: \(x^2+2x+3\ge0\)
\(x^2+6x+1=\left(2x+1\right).\sqrt{x^2+2x+3}\)
\(\Leftrightarrow x^2+2x+3+4x+2=\left(2x+1\right).\sqrt{x^2+2x+3}+4\)
Đặt \(a=\sqrt{x^2+2x+3}\); \(b=2x+1\), pt trở thành:
\(a^2+2b=ab+4\)
\(\Leftrightarrow a^2-4-ab+2b=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+2\right)-b\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a-b+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a-b=-2\end{matrix}\right.\)
.Với \(a=2\Leftrightarrow\sqrt{x^2+2x+3}=2\Leftrightarrow x^2+2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(N\right)\\x=-\sqrt{2}-1\left(N\right)\end{matrix}\right.\)
.Với \(a-b=-2\Leftrightarrow\sqrt{x^2+2x+3}-\left(2x+1\right)=-2\)
\(\Leftrightarrow\sqrt{x^2+2x+3}=-2+2x+1=2x-1\)
\(\Leftrightarrow x^2+2x+3=4x^2-4x+1\)
\(\Leftrightarrow3x^2-6x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{15}}{3}\left(N\right)\\x=\frac{3-\sqrt{15}}{3}\left(L\right)\end{matrix}\right.\)
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