+ \(P=x-2x^2=-2\left(x^2-\frac{1}{2}x\right)=-2\left(x^2-2x\cdot\frac{1}{4}+\frac{1}{16}-\frac{1}{16}\right)\)
\(=-2\left(x-\frac{1}{4}\right)^2+\frac{1}{8}\le\frac{1}{8}\forall x\)
Dấu "=" \(\Leftrightarrow-2\left(x-\frac{1}{4}\right)^2=0\Leftrightarrow x=\frac{1}{4}\)
Vậy Max \(P=\frac{1}{8}\Leftrightarrow x=\frac{1}{4}\)
+ \(Q=-\left[\left(x^2-2xy+y^2\right)+\left(y^2+2y\cdot\frac{1}{2}+\frac{1}{4}\right)-\frac{5}{4}\right]\)
\(=-\left[\left(x-y\right)^2+\left(y+\frac{1}{2}\right)^2\right]+\frac{5}{4}\le\frac{5}{4}\forall x,y\)
Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y+\frac{1}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\frac{1}{2}\)
Max \(Q=\frac{5}{4}\Leftrightarrow x=y=-\frac{1}{2}\)
