\(P=3x^2+y^2-2xy-3x+2\)
\(=x^2-2xy+y^2+2x^2-3x+2\)
\(=\left(x-y\right)^2+2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\)
do\(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-\frac{3}{4}\right)^2\ge0\end{cases}\Rightarrow P\ge\frac{7}{8}}\)
\(\Rightarrow P_{min}=\frac{7}{8}\)đạt được khi \(x=y=\frac{3}{4}\)