\(\left\{ \begin{array}{l}\dfrac{xy}{x+y}=\dfrac{(x-5)(y+10)}{x+y+5}\\\dfrac{xy}{x+y}=\dfrac{(x+40)(y-8)}{x+y+32}\end{array} \right.\) Tìm x và y

0 bình luận về “\(\left\{ \begin{array}{l}\dfrac{xy}{x+y}=\dfrac{(x-5)(y+10)}{x+y+5}\\\dfrac{xy}{x+y}=\dfrac{(x+40)(y-8)}{x+y+32}\end{array} \right.\) Tìm x và y”

1. Giải thích các bước giải:

    ĐKXĐ: $\left( {x + y} \right) \ne \left\{ {0; – 5; – 32} \right\}$

   Ta có:

   $\begin{array}{l}
   \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = \dfrac{{\left( {x – 5} \right)\left( {y + 10} \right)}}{{x + y + 5}}\\
   \dfrac{{xy}}{{x + y}} = \dfrac{{\left( {x + 40} \right)\left( {y – 8} \right)}}{{x + y + 32}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = \dfrac{{xy + 10x – 5y – 50}}{{x + y + 5}}\\
   \dfrac{{xy}}{{x + y}} = \dfrac{{xy – 8x + 40y – 320}}{{x + y + 32}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = \dfrac{{xy + 10x – 5y – 50}}{{x + y + 5}} = \dfrac{{xy – \left( {xy + 10x – 5y – 50} \right)}}{{x + y – \left( {x + y + 5} \right)}}\\
   \dfrac{{xy}}{{x + y}} = \dfrac{{xy – 8x + 40y – 320}}{{x + y + 32}} = \dfrac{{xy – \left( {xy – 8x + 40y – 320} \right)}}{{x + y – \left( {x + y + 32} \right)}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = \dfrac{{ – 10x + 5y + 50}}{{ – 5}}\\
   \dfrac{{xy}}{{x + y}} = \dfrac{{8x – 40y + 320}}{{ – 32}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = 2x – y – 10\\
   \dfrac{{xy}}{{x + y}} = \dfrac{{ – 1}}{4}x + \dfrac{5}{4}y – 10
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = 2x – y – 10\\
   2x – y – 10 = \dfrac{{ – 1}}{4}x + \dfrac{5}{4}y – 10
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{xy}}{{x + y}} = 2x – y – 10\\
   x = y
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{{x^2}}}{{2x}} = 2x – x – 10\\
   x = y
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{x}{2} = x – 10\\
   x = y
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   x = 20\\
   x = y
   \end{array} \right.\\
    \Leftrightarrow x = y = 20
   \end{array}$

   Vậy hệ có nghiệm duy nhất là: $(x;y)=(20;20)$

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