Có : \(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)\(\ge\left|x^2-x+1-x^2+x-2\right|=\left|-1\right|=1\)
Vậy Pmin=1\(\Leftrightarrow\left(x^2-x+1\right)\left(-x^2+x-2\right)\ge0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2-x+2\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x+1\ge0\\x^2-x+2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x+1\le0\\x^2-x+2\ge0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\in R\\x\in\varnothing\end{matrix}\right.\\\left\{{}\begin{matrix}x\in\varnothing\\x\in R\end{matrix}\right.\end{matrix}\right.\)
Vậy không tồn tại GTNN của P.
\(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)
\(P=\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right|+\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{7}{4}\right|\)
\(P=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right|+\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)
\(P=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{10}{4}\ge\dfrac{10}{4}=\dfrac{5}{2}\)
\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi \(x=\dfrac{1}{2}\)
