`(x+y)/3=(1+3y)/(2x)=(1+5y)/x `. tim x,y

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1. Đáp án:

   $\left( {x;y} \right) \in \left\{ {\left( {1;\dfrac{{ – 1}}{7}} \right);\left( {\dfrac{{ – 6}}{7};\dfrac{{ – 1}}{7}} \right)} \right\}$

   Giải thích các bước giải:

    ĐKXĐ:$x \ne 0$ 

   Ta có;

   $\dfrac{{x + y}}{3} = \dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 5y}}{x}\left( 1 \right)$

   Từ $(1)$ ta có:

   $\begin{array}{l}
   \dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 5y}}{x}\\
    \Leftrightarrow \dfrac{{1 + 3y}}{{2x}} = \dfrac{{2 + 10y}}{{2x}}\\
    \Leftrightarrow \dfrac{{2 + 10y}}{{2x}} – \dfrac{{1 + 3y}}{{2x}} = 0\\
    \Leftrightarrow \dfrac{{1 + 7y}}{{2x}} = 0\\
    \Leftrightarrow 1 + 7y = 0\\
    \Leftrightarrow y = \dfrac{{ – 1}}{7}
   \end{array}$

   Khi đó thay vào $(1)$ ta có: $\dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 3.\left( {\dfrac{{ – 1}}{7}} \right)}}{{2x}} = \dfrac{2}{{7x}}$

   Và:

   $\begin{array}{l}
   \dfrac{{x + \left( {\dfrac{{ – 1}}{7}} \right)}}{3} = \dfrac{2}{{7x}}\\
    \Leftrightarrow \dfrac{{7x – 1}}{{21}} = \dfrac{2}{{7x}}\\
    \Leftrightarrow \dfrac{{7x – 1}}{{21}} – \dfrac{2}{{7x}} = 0\\
    \Leftrightarrow \dfrac{{x\left( {7x – 1} \right) – 2.3}}{{21x}} = 0\\
    \Leftrightarrow \dfrac{{7{x^2} – x – 6}}{{21x}} = 0\\
    \Rightarrow 7{x^2} – x – 6 = 0\\
    \Leftrightarrow \left( {7{x^2} – 7x} \right) + \left( {6x – 6} \right) = 0\\
    \Leftrightarrow \left( {x – 1} \right)\left( {7x + 6} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   x = 1\\
   x = \dfrac{{ – 6}}{7}
   \end{array} \right.\left( {tm} \right)
   \end{array}$

   Vậy $\left( {x;y} \right) \in \left\{ {\left( {1;\dfrac{{ – 1}}{7}} \right);\left( {\dfrac{{ – 6}}{7};\dfrac{{ – 1}}{7}} \right)} \right\}$ thỏa mãn

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