Đáp án:
1. $\left( {x;y} \right) \in \left\{ {\left( {\dfrac{1}{2};\dfrac{{ - 1}}{{10}}} \right),\left( {\dfrac{{ - 1}}{2};\dfrac{1}{{10}}} \right)} \right\}$
Giải thích các bước giải:
1.Ta có:
$\begin{array}{l}
x\left( {x - y} \right) = \dfrac{3}{{10}}\left( 1 \right);y\left( {x - y} \right) = \dfrac{{ - 3}}{{50}}\left( 2 \right)\\
\Rightarrow x\left( {x - y} \right) - y\left( {x - y} \right) = \dfrac{3}{{10}} + \dfrac{3}{{50}}\\
\Leftrightarrow {\left( {x - y} \right)^2} = \dfrac{9}{{25}}\\
\Leftrightarrow \left[ \begin{array}{l}
x - y = \dfrac{3}{5}\\
x - y = \dfrac{{ - 3}}{5}
\end{array} \right.\\
\left( 1 \right) \Rightarrow x = \dfrac{1}{2} \Rightarrow y = \dfrac{{ - 1}}{{10}}\\
\left( 2 \right) \Rightarrow x = \dfrac{{ - 1}}{2} \Rightarrow y = \dfrac{1}{{10}}
\end{array}$
Vậy $\left( {x;y} \right) \in \left\{ {\left( {\dfrac{1}{2};\dfrac{{ - 1}}{{10}}} \right),\left( {\dfrac{{ - 1}}{2};\dfrac{1}{{10}}} \right)} \right\}$
2. Ta có:
$x + y = 2$
Lại có:
$\begin{array}{l}
{\left( {x - y} \right)^2} \ge 0,\dforall x,y \Rightarrow {x^2} + {y^2} \ge 2xy,\dforall x,y\\
\Rightarrow {\left( {x + y} \right)^2} \ge 4xy\\
\Rightarrow {2^2} \ge 4xy\\
\Rightarrow xy \le 1
\end{array}$
Dấu bằng xảy ra
$\left\{ \begin{array}{l}
x = y\\
x + y = 2
\end{array} \right. \Leftrightarrow x = y = 1$
Vậy ta có đpcm.
