\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=5-\left(x+1\right)^2\)
Ta có:
\(\left\{{}\begin{matrix}\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{4}+\sqrt{9}=5\\5-\left(x+1\right)^2\le5\end{matrix}\right.\)
Đẳng thức xảy ra khi và chỉ khi:
\(x+1=0\Leftrightarrow x=-1\)