1+7y/8=1+6y/11=1+9y/2x Tìm x,y

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1+7y/8=1+6y/11=1+9y/2x
Tìm x,y

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Ben Gia 6 years 2020-10-25T07:11:16+00:00 1 Answers 123 views 0

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  1. Calantha
    0
    2020-10-25T07:12:36+00:00
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    Đáp án:

    \(\left\{ \begin{array}{l}
    y =  – \dfrac{3}{{29}}\\
    x = 1
    \end{array} \right.\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    DK:x \ne 0\\
    \left\{ \begin{array}{l}
    \dfrac{{1 + 7y}}{8} = \dfrac{{1 + 6y}}{{11}}\\
    \dfrac{{1 + 6y}}{{11}} = \dfrac{{1 + 9y}}{{2x}}
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    11 + 77y = 8 + 48y\\
    2x + 12xy = 11 + 99y
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    29y =  – 3\\
    2x + 12xy = 11 + 99y
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    y =  – \dfrac{3}{{29}}\\
    2x + 12x.\left( { – \dfrac{3}{{29}}} \right) = 11 + 99.\left( { – \dfrac{3}{{29}}} \right)
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    y =  – \dfrac{3}{{29}}\\
    2x – \dfrac{{36}}{{29}}x = \dfrac{{22}}{{29}}
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    y =  – \dfrac{3}{{29}}\\
    \left( {2 – \dfrac{{36}}{{29}}} \right)x = \dfrac{{22}}{{29}}
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    y =  – \dfrac{3}{{29}}\\
    x = 1
    \end{array} \right.
    \end{array}\)

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