Giúp mình câu 208 Cảm ơn

Giúp mình câu 208 Cảm ơn

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  1. Đáp án:

    Giải thích các bước giải:

    \[\begin{array}{l}
    Neu\,m = 0\,thi\,y = \frac{{2x – 1}}{{\left( { – 2x + 1} \right)\left( {4{x^2} + 1} \right)}} = \frac{{ – 1}}{{4{x^2} + 1}}\\
    Do\,thi\,ham\,so\,luon\,co\,1TCN\,la\,y = 0\left( {TM} \right)\\
    Neu\,m \ne 0,\,de\,thay\,\mathop {\lim }\limits_{x \to \infty } y = 0 \Rightarrow TCN:y = 0\\
    De\,do\,thi\,ham\,so\,co\,dung\,1\,tiem\,can\,thi\,pt:\\
    \left( {m{x^2} – 2x + 1} \right)\left( {4{x^2} + 4m + 1} \right) = 0\,vo\,nghiem\\
    \Leftrightarrow \left\{ \begin{array}{l}
    m{x^2} – 2x + 1 = 0\,vo\,nghiem\\
    4{x^2} + 4m + 1 = 0\,vo\,nghiem
    \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
    \Delta ‘ = 1 – m < 0\\ \Delta ' = - 4m - 1 < 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m > 1\\
    m > \frac{{ – 1}}{4}
    \end{array} \right. \Leftrightarrow m > 1\\
    Vay\,\left[ \begin{array}{l}
    m > 1\\
    m \ne 0
    \end{array} \right.\,hay\,m \in \left\{ 0 \right\} \cup \left( {1; + \infty } \right)
    \end{array}\]

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