ĐK:y\(\ge0\)
\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1=\left(x^2-x\sqrt{y}+\dfrac{y}{4}+x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{4}\right)+\left(\dfrac{3}{4}y-\dfrac{\sqrt{y}}{2}+\dfrac{1}{12}\right)+\dfrac{2}{3}=\left(x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}y-\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\in R;y\ge0\)
=>Min P=\(\dfrac{2}{3}\)đạt được khi \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}=0\\\dfrac{\sqrt{3}}{2}y-\dfrac{\sqrt{3}}{6}=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{3}-3}{6}\\y=\dfrac{1}{3}\end{matrix}\right.\)
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ĐK:y\(\ge0\)
\(P=x^2-x\sqrt{y}+x+y-\sqrt{y}+1=\left(x^2-x\sqrt{y}+\dfrac{y}{4}-\dfrac{\sqrt{y}}{2}+\dfrac{1}{4}+x\right)+\left(\dfrac{3}{4}y-\dfrac{\sqrt{y}}{2}+\dfrac{1}{12}\right)+\dfrac{2}{3}=\left(x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3y}}{2}-\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\in R;y\ge0\)
=>Min P=\(\dfrac{2}{3}\)đạt được khi \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{y}}{2}+\dfrac{1}{2}=0\\\dfrac{\sqrt{3y}}{2}-\dfrac{\sqrt{3}}{6}=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{9}\end{matrix}\right.\)
