Đáp án:
\({P_{\min }} = 2\sqrt 2\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{{x^2} + {y^2}}}{{x - y}} = \dfrac{{{{\left( {x - y} \right)}^2} + 2xy}}{{x - y}} = x - y + \dfrac{2}{{x - y}} \ge 2\sqrt {\left( {x - y} \right).\dfrac{2}{{x - y}}} = 2\sqrt 2 \\
\Rightarrow {P_{\min }} = 2\sqrt 2 \Leftrightarrow x - y = \dfrac{2}{{x - y}}\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 2\\
xy = 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x - y = \sqrt 2 \\
xy = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - \sqrt 2 + \sqrt 6 }}{2}\\
y = \dfrac{2}{{ - \sqrt 2 + \sqrt 6 }}
\end{array} \right.
\end{array}\)
