\(x^2+y^2+xy+2-3x-3y=0\)
\(\Leftrightarrow\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2=1\)
Đặt \(\left\{{}\begin{matrix}x+\dfrac{y}{2}-\dfrac{3}{2}=sina\\\dfrac{\sqrt{3}}{2}\left(y-1\right)=cosa\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=sina-\dfrac{y-3}{2}\\y=\dfrac{2}{\sqrt{3}}cosa+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=sina-\dfrac{1}{\sqrt{3}}cosa+1\\y=\dfrac{2}{\sqrt{3}}cosa+1\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{3sina+\dfrac{1}{\sqrt{3}}cosa+6}{sina+\dfrac{1}{\sqrt{3}}cosa+8}\)
\(\Rightarrow P.sina-\dfrac{P}{\sqrt{3}}cosa+8P=3sina+\dfrac{1}{\sqrt{3}}cosa+6\)
\(\Rightarrow\left(P-3\right)sina-\left(\dfrac{P+1}{\sqrt{3}}\right)cosa=6-8P\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(P-3\right)^2+\left(\dfrac{P+1}{\sqrt{3}}\right)^3\ge\left(6-8P\right)^2\)
\(\Leftrightarrow47P^2-67P+20\le0\)
\(\Rightarrow\dfrac{20}{47}\le P\le1\)
