\(\Leftrightarrow x^4+2x^3+x^2+x^2+2x+1+\sqrt{x^2+2x+10}=3\)
\(\Leftrightarrow\left(x^2+x\right)^2+\left(x+1\right)^2+\sqrt{\left(x+1\right)^2+9}=3\)
Do \(\left\{{}\begin{matrix}\left(x^2+x\right)^2\ge0\\\left(x+1\right)^2\ge0\\\sqrt{\left(x+1\right)^2+9}\ge3\end{matrix}\right.\)
\(\Rightarrow VT\ge3\)
Dấu "=" xảy ra khi và chỉ khi \(x+1=0\Leftrightarrow x=-1\)