ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left(x+1\right)^3+\left(3x^2+6x+3-4\left(x+3\right)\right)\sqrt{x+3}=0\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{x+3}=b\end{matrix}\right.\)
\(\Rightarrow a^3+\left(3a^2-4b^2\right)b=0\)
\(\Leftrightarrow a^3+3a^2b-4b^3=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+2b\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2b=-a\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=x+1\left(x\ge-1\right)\\2\sqrt{x+3}=-x-1\left(x\le-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2+2x+1\\4\left(x+3\right)=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-2x-11=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=1-2\sqrt{3}\end{matrix}\right.\)
