1. tìm x,y :x(x -y)=3/10;y(x -y)=-3/50 2. x+y=2 cmr xy < hoặc = 1

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1. 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.

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