Giải:
\begin{array}{l} Do\,\left\{ \begin{array}{l} x,y \in R\\ {x^2} + {y^2} = 1 \end{array} \right. \Rightarrow Coi\,\left\{ \begin{array}{l} x = sin\,\alpha \\ y = cos\,\alpha \\ \alpha \in \left[ {0;2\pi } \right] \end{array} \right.\\ \Rightarrow P = \frac{{2\left( {si{n^2}\alpha + 6sin\alpha cos\alpha } \right)}}{{1 + 2sin\alpha cos\alpha + 2co{s^2}\alpha }} = \frac{{6\sin 2\alpha - cos2\alpha + 1}}{{\sin 2\alpha + cos2\alpha + 2}}\\ \Leftrightarrow P\left( {\sin 2\alpha + cos2\alpha + 2} \right) = 6\sin 2\alpha - cos2\alpha + 1\\ \Leftrightarrow \left( {P - 6} \right)\sin 2\alpha + \left( {P + 1} \right)cos2\alpha = 1 - 2P\\ \Rightarrow \exists \alpha \Leftrightarrow {\left( {P - 6} \right)^2} + {\left( {P + 1} \right)^2} \ge {\left( {1 - 2P} \right)^2} \Leftrightarrow {P^2} + 3P - 18 \le 0 \Leftrightarrow - 6 \le P \le 3\\ \bullet \,Min\,P = - 6 \Leftrightarrow 12\sin 2\alpha + 5cos2\alpha = 13 \Leftrightarrow \left\{ \begin{array}{l} t = \tan \alpha \\ \frac{{24t}}{{1 + {t^2}}} + \frac{{5\left( {1 - {t^2}} \right)}}{{1 + {t^2}}} = 13 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t = \tan \alpha \\ {\left( {3t - 2} \right)^2} = 0 \end{array} \right.\\ \Leftrightarrow t = \tan \alpha = \frac{x}{y} = \frac{2}{3} \Leftrightarrow \frac{{{x^2}}}{4} = \frac{{{y^2}}}{9} = \frac{{{x^2} + {y^2}}}{{13}} = \frac{1}{{13}} \Rightarrow \left\{ \begin{array}{l} x = \frac{{ \pm 2}}{{\sqrt {13} }}\\ y = \frac{{ \pm 3}}{{\sqrt {13} }} \end{array} \right.\\ \bullet MaxP = 3 \Leftrightarrow 3\sin 2\alpha - 4cos2\alpha = 5 \Leftrightarrow \left\{ \begin{array}{l} t = \tan \alpha \\ \frac{{6t}}{{1 + {t^2}}} - \frac{{4\left( {1 - {t^2}} \right)}}{{1 + {t^2}}} = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} t = \tan \alpha \\ {\left( {t - 3} \right)^2} = 0 \end{array} \right.\\ \Leftrightarrow t = \tan \alpha = \frac{x}{y} = 3 \Leftrightarrow \frac{{{x^2}}}{9} = \frac{{{y^2}}}{1} = \frac{{{x^2} + {y^2}}}{{10}} = \frac{1}{{10}} \Rightarrow \left\{ \begin{array}{l} x = \frac{{ \pm 3}}{{\sqrt {10} }}\\ y = \frac{{ \pm 1}}{{\sqrt {13} }} \end{array} \right. \end{array}
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