Đáp án:
$\begin{array}{l}
1)\\
A = \dfrac{{{x^2} + {y^2}}}{{x - y}} = \dfrac{{{x^2} - 2xy + {y^2} + 2xy}}{{x - y}}\\
= \dfrac{{{{\left( {x - y} \right)}^2} + 2}}{{x - y}}\left( {do:x.y = 1} \right)\\
= x - y + \dfrac{2}{{x - y}}\\
Do:x > y \Rightarrow x - y > 0\\
Theo\,CO - si:\\
\left( {x - y} \right) + \dfrac{2}{{x - y}} \ge 2\sqrt {\left( {x - y} \right).\dfrac{2}{{x - y}}} \\
\Rightarrow A \ge 2\sqrt 2 \\
\Rightarrow GTNN:A = 2\sqrt 2 \\
Khi:\left( {x - y} \right) = \dfrac{2}{{x - y}}\\
\Rightarrow {\left( {x - y} \right)^2} = 2\\
\Rightarrow x - y = \sqrt 2 \\
\Rightarrow x = y + \sqrt 2 \\
\Rightarrow y.\left( {y + \sqrt 2 } \right) = 1\\
\Rightarrow {y^2} + \sqrt 2 .y - 1 = 0\\
\Rightarrow \left[ \begin{array}{l}
y = \dfrac{{\sqrt 6 - \sqrt 2 }}{2} \Rightarrow x = \dfrac{{\sqrt 6 + \sqrt 2 }}{2}\\
y = \dfrac{{ - \sqrt 6 - \sqrt 2 }}{2} \Rightarrow x = \dfrac{{ - \sqrt 6 + \sqrt 2 }}{2}
\end{array} \right.\\
2)\\
M = {x^3} + {y^3} = \left( {x + y} \right).\left( {{x^2} - xy + {y^2}} \right)\\
= 1.\left( {{x^2} + 2xy + {y^2} - 3xy} \right)\\
= {\left( {x + y} \right)^2} - 3xy\\
= 1 - 3xy\left( {do:x + y = 1} \right)\\
Do:x + y = 1 \ge 2\sqrt {xy} \\
\Rightarrow \sqrt {xy} \le \dfrac{1}{2}\\
\Rightarrow x.y \le \dfrac{1}{4}\\
\Rightarrow 3xy \le \dfrac{3}{4}\\
\Rightarrow - 3xy \ge - \dfrac{3}{4}\\
\Rightarrow 1 - 3xy \ge 1 - \dfrac{3}{4}\\
\Rightarrow M \ge \dfrac{1}{4}\\
\Rightarrow GTNN:M = \dfrac{1}{4}\\
Khi:x = y = \dfrac{1}{2}
\end{array}$
