Pt \(\sqrt{3x^2+6x+4}+\sqrt{2x^2+4x+11}=\left(x+3\right)\left(1-x\right)\left(1\right)\)
VT=\(\sqrt{3\left(x+1\right)^2+1}+\sqrt{2\left(x+1\right)^2+9}\ge\sqrt{1}+\sqrt{9}=4\)
\(VP=\left(x+3\right)\left(1-x\right)\le\frac{1}{4}\left(x+3+1-x\right)^2=4\)
Khi đó (1) xảy ra khi \(\left\{{}\begin{matrix}x+1=0\\x+3=1-x\end{matrix}\right.\)=> \(x=-1\)
Vậy x=-1