Tìm x,y,z biết a, $\frac{6}{11}$ .x=$\frac{9}{2}$ .y=$\frac{11}{5}$ .z và x+y+z=240 b, $\frac{x-y}{3}$= $\frac{x+y}{15}$= $\frac{x.y}{200}$

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1. Đáp án:

   $a)\left( {x;y;z} \right) = \left( {\dfrac{{87120}}{{497}};\dfrac{{10560}}{{497}};\dfrac{{21600}}{{497}}} \right)$

   $b)\left( {x;y} \right) \in \left\{ {\left( {0;0} \right),\left( {\dfrac{{100}}{3};\dfrac{{200}}{9}} \right)} \right\}$

   Giải thích các bước giải:

    a) Ta có:

   $\begin{array}{l}
   \dfrac{6}{{11}}x = \dfrac{9}{2}y = \dfrac{{11}}{5}z\\
    \Rightarrow \dfrac{x}{{\dfrac{{11}}{6}}} = \dfrac{y}{{\dfrac{2}{9}}} = \dfrac{z}{{\dfrac{5}{{11}}}}\\
    \Rightarrow \dfrac{x}{{\dfrac{{11}}{6}}} = \dfrac{y}{{\dfrac{2}{9}}} = \dfrac{z}{{\dfrac{5}{{11}}}} = \dfrac{{x + y + z}}{{\dfrac{{11}}{6} + \dfrac{2}{9} + \dfrac{5}{{11}}}} = \dfrac{{240}}{{\dfrac{{497}}{{198}}}} = \dfrac{{47520}}{{497}}\\
    \Rightarrow \left\{ \begin{array}{l}
   x = \dfrac{{11}}{6}.\dfrac{{47520}}{{497}} = \dfrac{{87120}}{{497}}\\
   y = \dfrac{2}{9}.\dfrac{{47520}}{{497}} = \dfrac{{10560}}{{497}}\\
   z = \dfrac{5}{{11}}.\dfrac{{47520}}{{497}} = \dfrac{{21600}}{{497}}
   \end{array} \right.
   \end{array}$

   $ \Rightarrow \left( {x;y;z} \right) = \left( {\dfrac{{87120}}{{497}};\dfrac{{10560}}{{497}};\dfrac{{21600}}{{497}}} \right)$

   Vậy $\left( {x;y;z} \right) = \left( {\dfrac{{87120}}{{497}};\dfrac{{10560}}{{497}};\dfrac{{21600}}{{497}}} \right)$

   b) Ta có:

   $\dfrac{{x – y}}{3} = \dfrac{{x + y}}{{15}} = \dfrac{{xy}}{{200}}\left( 1 \right)$

   Lại có:

   $\begin{array}{l}
   \left( 1 \right) \Rightarrow \dfrac{{x – y}}{3} = \dfrac{{x + y}}{{15}} = \dfrac{{x – y + x + y}}{{3 + 15}} = \dfrac{{2x}}{{18}} = \dfrac{x}{9}\\
   \left( 1 \right) \Rightarrow \dfrac{{x – y}}{3} = \dfrac{{x + y}}{{15}} = \dfrac{{x + y – \left( {x – y} \right)}}{{15 – 3}} = \dfrac{{2y}}{{12}} = \dfrac{y}{6}
   \end{array}$

   Như vậy: $\left( 1 \right) \Rightarrow \dfrac{x}{9} = \dfrac{y}{6} = \dfrac{{xy}}{{200}}\left( 2 \right)$

   Khi đó:

   $\begin{array}{l}
    \Rightarrow \dfrac{x}{9} = \dfrac{{xy}}{{200}}\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   \dfrac{y}{{200}} = \dfrac{1}{9}
   \end{array} \right.\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   y = \dfrac{{200}}{9}
   \end{array} \right.
   \end{array}$

   $\begin{array}{l}
    + )TH1:x = 0\\
   \left( 2 \right) \Rightarrow y = \dfrac{x}{9}.6 = 0\\
    \Rightarrow \left( {x;y} \right) = \left( {0;0} \right)
   \end{array}$

   $\begin{array}{l}
    + )TH2:y = \dfrac{{200}}{9}\\
   \left( 2 \right) \Rightarrow x = 9.\dfrac{y}{6} = \dfrac{3}{2}.\dfrac{{200}}{9} = \dfrac{{100}}{3}\\
    \Rightarrow \left( {x;y} \right) = \left( {\dfrac{{100}}{3};\dfrac{{200}}{9}} \right)
   \end{array}$

   Vậy $\left( {x;y} \right) \in \left\{ {\left( {0;0} \right),\left( {\dfrac{{100}}{3};\dfrac{{200}}{9}} \right)} \right\}$

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