giúp mikkk vsss :))))

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giúp mikkk vsss :))))
giup-mikkk-vsss

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Xavia 5 years 2021-05-24T04:53:15+00:00 1 Answers 20 views 0

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  1. binhan
    0
    2021-05-24T04:54:47+00:00
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    Đáp án:

    $m \in \left\{ { – 5;\dfrac{7}{5}} \right\}$

    Giải thích các bước giải:

    c) Ta có:

    Phương trình ${x^2} – 2\left( {m – 1} \right)x + m – 3 = 0$ có 2 nghiệm $x_1;x_2$

    Theo ĐL Viet ta có:

    $\left\{ \begin{array}{l}
    {x_1} + {x_2} = 2\left( {m – 1} \right)\\
    {x_1}{x_2} = m – 3
    \end{array} \right.$

    Để $\left| {\dfrac{1}{{{x_1}}} – \dfrac{1}{{{x_2}}}} \right|$ tồn tại 

    $ \Leftrightarrow {x_1};{x_2} \ne 0 \Rightarrow {x_1}{x_2} \ne 0 \Rightarrow m – 3 \ne 0 \Rightarrow m \ne 3$

    Khi đó:

    $\begin{array}{l}
    \left| {\dfrac{1}{{{x_1}}} – \dfrac{1}{{{x_2}}}} \right| = \dfrac{{\sqrt {11} }}{2}\\
     \Leftrightarrow {\left( {\dfrac{1}{{{x_1}}} – \dfrac{1}{{{x_2}}}} \right)^2} = \dfrac{{11}}{4}\\
     \Leftrightarrow \dfrac{{{{\left( {{x_2} – {x_1}} \right)}^2}}}{{{{\left( {{x_1}{x_2}} \right)}^2}}} = \dfrac{{11}}{4}\\
     \Leftrightarrow 4{\left( {{x_2} – {x_1}} \right)^2} – 11{\left( {{x_1}{x_2}} \right)^2} = 0\\
     \Leftrightarrow 4\left( {{{\left( {{x_1} + {x_2}} \right)}^2} – 4{x_1}{x_2}} \right) – 11{\left( {{x_1}{x_2}} \right)^2} = 0\\
     \Leftrightarrow 4\left( {4{{\left( {m – 1} \right)}^2} – 4\left( {m – 3} \right)} \right) – 11{\left( {m – 3} \right)^2} = 0\\
     \Leftrightarrow 5{m^2} + 18m – 35 = 0\\
     \Leftrightarrow \left( {m + 5} \right)\left( {5m – 7} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    m =  – 5\\
    m = \dfrac{7}{5}
    \end{array} \right.\left( {tm} \right)
    \end{array}$

    Vậy $m \in \left\{ { – 5;\dfrac{7}{5}} \right\}$

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )