`~rai~`
\(2x^4+2x^3+5x^2+2x+2=0\\\Leftrightarrow (x^4+2x^3+x^2)+(x^4+2x^2+1)+\left(2x^2+2x+\dfrac{1}{2}\right)+\dfrac{1}{2}=0\\\Leftrightarrow (x^2+x)^2+(x^2+1)+2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0.\quad(1)\\\text{Ta có:}(x^2+x)\ge 0;(x^2+1)\ge 0;2\left(x+\dfrac{1}{2}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow (x^2+x)+(x^2+1)+2\left(x+\dfrac{1}{2}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow (x^2+x)^2+(x^2+1)^2+2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge \dfrac{1}{2}>0\quad\forall x\in\mathbb{R}\\\Rightarrow \text{(1) vô nghiệm.}\\\text{Vậy phương trình vô nghiệm.}\)