giải bpt (x^2+1)/ x^2+3x-10 <0 (x-5) nhân căn -x^2+7x-6 <= 0

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giải bpt (x^2+1)/ x^2+3x-10 <0 (x-5) nhân căn -x^2+7x-6 <= 0

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Parker 2 tuần 2021-12-03T22:23:13+00:00 1 Answers 2 views 0

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    0
    2021-12-03T22:24:39+00:00

    Đáp án:

     a. \( – 5 < x < 2\)

    Giải thích các bước giải:

    \(\begin{array}{l}
    a.\frac{{{x^2} + 1}}{{{x^2} + 3x – 10}} < 0\\
    Do:{x^2} + 1 > 0\forall x \in R\\
     \to {x^2} + 3x – 10 < 0\\
     \to \left( {x – 2} \right)\left( {x + 5} \right) < 0\\
     \to \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x – 2 > 0\\
    x + 5 < 0
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x – 2 < 0\\
    x + 5 > 0
    \end{array} \right.
    \end{array} \right. \to \left[ \begin{array}{l}
    (loai)\\
     – 5 < x < 2
    \end{array} \right.\\
    b.\left( {x – 5} \right)\left( {\sqrt { – {x^2} + 7x – 6} } \right) \le 0\\
     \to \left\{ \begin{array}{l}
     – {x^2} + 7x – 6 \ge 0\\
    x – 5 \le 0
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    \left( {x – 6} \right)\left( {x – 1} \right) \le 0\\
    x \le 5
    \end{array} \right.\\
     \to \left\{ \begin{array}{l}
    x \in \left[ {1;6} \right]\\
    x \le 5
    \end{array} \right.\\
    KL:x \in \left[ {1;5} \right]
    \end{array}\)

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