$\begin{array}{l}
1 - {y^2} = x\left( {x - y} \right)\\
\leftrightarrow {x^2} + {y^2} = 1 + xy\\
\cos i:\\
{x^2} + {y^2} \ge 2\sqrt {{x^2}{y^2}} = 2\left| {xy} \right|\\
\to 1 + xy = {x^2} + {y^2} \ge 2\left| {xy} \right| \ge 2xy \leftrightarrow xy \le 1\;dau = \leftrightarrow \left\{ \begin{array}{l}
{x^2} = {y^2}\\
xy = 1
\end{array} \right.\\
1 + xy \ge 2\left| {xy} \right| \ge - 2xy \leftrightarrow xy \ge \frac{{ - 1}}{3}\;dau = \leftrightarrow \left\{ \begin{array}{l}
{x^2} = {y^2}\\
xy = - \frac{1}{3}
\end{array} \right.\\
\to - \frac{1}{3} \le xy \le 1\\
P = \frac{{{x^6} + {y^6} - 1}}{{{x^3}y + x{y^3}}} = \frac{{{{\left( {{x^2} + {y^2}} \right)}^3} - 3{x^2}{y^2}\left( {{x^2} + {y^2}} \right) - 1}}{{xy\left( {{x^2} + {y^2}} \right)}} = \frac{{{{\left( {1 + xy} \right)}^3} - 3{x^2}{y^2}\left( {1 + xy} \right) - 1}}{{xy\left( {1 + xy} \right)}} = \frac{{ - 2{x^3}{y^3} + 3xy}}{{xy\left( {1 + xy} \right)}}\\
dat\;t = xy \to - \frac{1}{3} \le t \le 1\\
P = \frac{{ - 2{t^3} + 3t}}{{t\left( {1 + t} \right)}}\\
xet\;P(t)\;tren\;doan\;\left[ { - \frac{1}{3};1} \right]\\
P\;xac\;dinh \leftrightarrow t \ne 0\;( - 1\;khong\;thuoc\;doan\;dang\;xet\;nhe)\\
ta\;co\;\mathop {\lim }\limits_{t \to 0} P = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - 2{t^3} + 3t}}{{t\left( {1 + t} \right)}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - 2{t^2} + 3}}{{1 + t}}} \right) = 3\\
khi\;t \ne 0\; \to P(t) = \frac{{ - 2{t^2} + 3}}{{1 + t}}\\
{P^/} = \frac{{ - 2{t^2} - 4t - 3}}{{{{\left( {1 + t} \right)}^2}}} < 0\;\forall t\\
co\;P\left( { - \frac{1}{3}} \right) = \frac{{25}}{6}\\
P\left( 1 \right) = \frac{1}{2}\\
\to \frac{1}{2} < 3 < \frac{{25}}{6}\\
vay\;minP = \frac{1}{2}\;dau = \leftrightarrow t = 1 \leftrightarrow x,y = ...\\
\max P = \frac{{25}}{6}{\rm{\backslash dau = }} \leftrightarrow t = - \frac{1}{3} \leftrightarrow x,y = ...
\end{array}$
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