Giúp mk vs nhé!!!!!!!

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Giúp mk vs nhé!!!!!!!

giup-mk-vs-nhe

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Sapo 1 year 2020-11-27T08:32:31+00:00 1 Answers 47 views 0

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  1. Neala
    0
    2020-11-27T08:34:16+00:00
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    Đáp án:

    $\begin{array}{l}
    a)\dfrac{{x + 5}}{{2017}} – \dfrac{{x + 5}}{{2018}} + \dfrac{{x + 5}}{{2019}} – \dfrac{{x + 5}}{{2020}} = 0\\
     \Rightarrow \left( {x + 5} \right).\left( {\dfrac{1}{{2017}} – \dfrac{1}{{2018}} + \dfrac{1}{{2019}} – \dfrac{1}{{2020}}} \right) = 0\\
     \Rightarrow x + 5 = 0\\
     \Rightarrow x =  – 5\\
    Vay\,x =  – 5\\
    b)\dfrac{{x + 13}}{{2006}} + \dfrac{{x + 2006}}{{13}} + \dfrac{{x + 1}}{{2018}} + 3 = 0\\
     \Rightarrow \dfrac{{x + 13}}{{2006}} + 1 + \dfrac{{x + 2006}}{{13}} + 1 + \dfrac{{x + 1}}{{2018}} + 1 = 0\\
     \Rightarrow \dfrac{{x + 13 + 2006}}{{2006}} + \dfrac{{x + 2006 + 13}}{{13}} + \dfrac{{x + 1 + 2018}}{{2018}} = 0\\
     \Rightarrow \dfrac{{x + 2019}}{{2006}} + \dfrac{{x + 2019}}{{13}} + \dfrac{{x + 2019}}{{2018}} = 0\\
     \Rightarrow \left( {x + 2019} \right)\left( {\dfrac{1}{{2006}} + \dfrac{1}{{13}} + \dfrac{1}{{2018}}} \right) = 0\\
     \Rightarrow x + 2019 = 0\\
     \Rightarrow x =  – 2019\\
    Vay\,x =  – 2019\\
    c){x^2} – 3x > 0\\
     \Rightarrow x\left( {x – 3} \right) > 0\\
     \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x > 0\\
    x – 3 > 0
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x < 0\\
    x – 3 < 0
    \end{array} \right.
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x > 0\\
    x > 3
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x < 0\\
    x < 3
    \end{array} \right.
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x > 3\\
    x < 0
    \end{array} \right.\\
    Vay\,x > 3\,hoac\,x < 0\\
    d)Dkxd:\left\{ \begin{array}{l}
    x \ne  – 3\\
    x \ne  – 7\\
    x \ne  – 10
    \end{array} \right.\\
    \dfrac{4}{{\left( {x + 3} \right)\left( {x + 7} \right)}} + \dfrac{3}{{\left( {x + 7} \right)\left( {x + 10} \right)}}\\
     = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow \dfrac{{\left( {x + 7} \right) – \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 7} \right)}} + \dfrac{{\left( {x + 10} \right) – \left( {x + 7} \right)}}{{\left( {x + 7} \right)\left( {x + 10} \right)}}\\
     = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow \dfrac{1}{{x + 3}} – \dfrac{1}{{x + 7}} + \dfrac{1}{{x + 7}} – \dfrac{1}{{x + 10}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow \dfrac{1}{{x + 3}} – \dfrac{1}{{x + 10}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow \dfrac{{x + 10 – \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x + 10} \right)}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow \dfrac{7}{{\left( {x + 3} \right)\left( {x + 10} \right)}} = \dfrac{x}{{\left( {x + 3} \right)\left( {x + 10} \right)}}\\
     \Rightarrow x = 7\left( {tmdk} \right)\\
    Vay\,x = 7
    \end{array}$

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