Ta có: P= \(x^2+xy+y^2-2x-3y+2010\)
\(\Leftrightarrow\) 4P= \(4\left(x^2+xy+y^2-2x-3y+2010\right)\)
= \(4x^2+4xy+4y^2-8x-12y+8040\)
= \(\left(4x^2+y^2+4+4xy-8x-8y\right)+3y^2-8y+8036\)
= \(\left(2x+y-2\right)^2+3y^2-8y+\dfrac{16}{3}-\dfrac{16}{3}+8036\)
= \(\left(2x+y-2\right)^2+3\left(y^2-\dfrac{8}{3}y+\dfrac{16}{9}\right)+\dfrac{24092}{3}\)
= \(\left(2x+y-2\right)^2+3\left(y-\dfrac{4}{3}\right)^2+\dfrac{24092}{3}\) \(\geq\) \(\dfrac{24092}{3}\)
\(\Rightarrow\) 4P \(\geq\) \(\dfrac{24092}{3}\) \(\Rightarrow\) P \(\geq\) \(\dfrac{6023}{3}\)
Dấu = xảy ra khi \(\begin{cases} (2x+y-2)^{2}=0\\ (y-\dfrac{4}{3})^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x+y-2=0\\ y-\dfrac{4}{3}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(y-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} 2x=-(\dfrac{4}{3}-2)\\ y=\dfrac{4}{3} \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)
Từ đó suy ra Min P= \(\dfrac{6023}{3}\) khi \(\begin{cases} x=\dfrac{1}{3}\\ y=\dfrac{4}{3} \end{cases} \)
Chúc bạn học tốt. 
