Câu 12 aaaaaaaaaa……….

Câu 12 aaaaaaaaaa……….

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  1. Đáp án:

    Giải thích các bước giải:

    \[\begin{array}{l}
    \frac{{2{x^2} + \left( {1 – m} \right)x + 1 + m}}{{x – m}}\\
    y’ = \frac{{\left( {4x + 1 – m} \right)\left( {x – m} \right) – 2{x^2} – \left( {1 – m} \right)x – 1 – m}}{{{{\left( {x – m} \right)}^2}}}\\
    y’ = \frac{{2{x^2} – 4mx – m + {m^2} – 2m – 1}}{{{{\left( {x – m} \right)}^2}}}\\
    y’ = \frac{{2{x^2} – 4mx + {m^2} – 2m – 1}}{{{{\left( {x – m} \right)}^2}}}\\
    Ham\,so\,dong\,bien\,tren\,\left( {1; + \infty } \right) \Leftrightarrow y’ \ge 0\,\,\,\forall x \in \left( {1; + \infty } \right)\\
    \Leftrightarrow \left\{ \begin{array}{l}
    2{x^2} – 4mx + {m^2} – 2m – 1 \ge 0\,\,\,\forall x \in \left( {1; + \infty } \right)\\
    m \notin \left( {1; + \infty } \right)
    \end{array} \right.\\
    \Delta ‘ = 4{m^2} – 2\left( {{m^2} – 2m + 1} \right) = 2{m^2} + 4m + 2 = 2\left( {{m^2} + 2m + 1} \right) = 2{\left( {m + 1} \right)^2}\\
    \Rightarrow y’ = 0\,luon\,co\,2\,nghiem\,{x_1};{x_2} = \frac{{2m \pm \left( {m + 1} \right)\sqrt 2 }}{2}\\
    Bai\,toan\,thoa\, \Leftrightarrow \left\{ \begin{array}{l}
    {x_1} \le {x_2} \le 1\\
    m \notin \left( {1; + \infty } \right)
    \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
    \frac{{2m + \left( {m + 1} \right)\sqrt 2 }}{2} \le 1\\
    m \le 1
    \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
    2m + \left( {m + 1} \right)\sqrt 2 \le 2\\
    m \le 1
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    \left( {2 + \sqrt 2 } \right)m \le 2 – \sqrt 2 \\
    m \le 1
    \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
    m \le \frac{{2 – \sqrt 2 }}{{2 + \sqrt 2 }}\\
    m \le 1
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    m \le \frac{{6 – 4\sqrt 2 }}{{\sqrt 2 }}\\
    m \le 1
    \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
    m \le 3 – 2\sqrt 2 \\
    m \le 1
    \end{array} \right. \Leftrightarrow m \le 3 – 2\sqrt 2
    \end{array}\]

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