Tìm x, biết: a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0

Question

Tìm x, biết:
a, $\frac{x-2}{x-6}$ < 0 b, $x^{2}$ + 4x > 0

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Vodka 1 year 2020-11-27T05:27:16+00:00 1 Answers 51 views 0

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    0
    2020-11-27T05:29:02+00:00

    Đáp án:

    $\begin{array}{l}
    a)\dfrac{{x – 2}}{{x – 6}} < 0\\
     \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x – 2 > 0\\
    x – 6 < 0
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x – 2 < 0\\
    x – 6 > 0
    \end{array} \right.
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x > 2\\
    x < 6
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x < 2\\
    x > 6
    \end{array} \right.\left( {ktm} \right)
    \end{array} \right.\\
     \Rightarrow 2 < x < 6\\
    \text{Vậy}\,2 < x < 6\\
    b){x^2} + 4x > 0\\
     \Rightarrow x\left( {x + 4} \right) > 0\\
     \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x > 0\\
    x + 4 > 0
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x < 0\\
    x + 4 < 0
    \end{array} \right.
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    \left\{ \begin{array}{l}
    x > 0\\
    x >  – 4
    \end{array} \right.\\
    \left\{ \begin{array}{l}
    x < 0\\
    x <  – 4
    \end{array} \right.
    \end{array} \right.\\
     \Rightarrow \left[ \begin{array}{l}
    x > 0\\
    x <  – 4
    \end{array} \right.\\
    \text{Vậy}\,x > 0\,\text{hoặc}\,x <  – 4
    \end{array}$

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