\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1\)
\(=\left(x^2-x\sqrt{y}+\dfrac{1}{4}y\right)+\left(x-\dfrac{1}{2}\sqrt{y}\right)+\dfrac{1}{4}+\left(\dfrac{3}{4}y-\dfrac{1}{2}\sqrt{y}+\dfrac{1}{12}\right)-\dfrac{1}{12}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\sqrt{y}\right)^2+\left(x-\dfrac{1}{2}\sqrt{y}\right)+\dfrac{1}{4}+\dfrac{3}{4}\left(y-\dfrac{2}{3}\sqrt{y}+\dfrac{1}{9}\right)+\dfrac{2}{3}\)
\(=\left(x-\dfrac{1}{2}\sqrt{y}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(\sqrt{y}-\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\) Đạt GTNN là \(\dfrac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}\sqrt{y}+\dfrac{1}{2}=0\\\sqrt{y}-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{9}\end{matrix}\right.\)