1/2-/5/4-2x/=1/3 2x-/x+1/=-1/2 /2x-1/-/x+1/3/=0

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1/2-/5/4-2x/=1/3
2x-/x+1/=-1/2
/2x-1/-/x+1/3/=0

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Amity 1 year 2020-11-28T12:23:39+00:00 1 Answers 84 views 0

Answers ( )

    -1
    2020-11-28T12:24:47+00:00

    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    a,\\
    \dfrac{1}{2} – \left| {\dfrac{5}{4} – 2x} \right| = \dfrac{1}{3}\\
     \Leftrightarrow \left| {\dfrac{5}{4} – 2x} \right| = \dfrac{1}{2} – \dfrac{1}{3}\\
     \Leftrightarrow \left| {\dfrac{5}{4} – 2x} \right| = \dfrac{1}{6}\\
     \Leftrightarrow \left[ \begin{array}{l}
    \dfrac{5}{4} – 2x = \dfrac{1}{6}\\
    \dfrac{5}{4} – 2x =  – \dfrac{1}{6}
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    2x = \dfrac{5}{4} – \dfrac{1}{6}\\
    2x = \dfrac{5}{4} + \dfrac{1}{6}
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    2x = \dfrac{{13}}{{12}}\\
    2x = \dfrac{{17}}{{12}}
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{{13}}{{24}}\\
    x = \dfrac{{17}}{{24}}
    \end{array} \right.\\
    b,\\
    2x – \left| {x + 1} \right| =  – \dfrac{1}{2}\\
     \Leftrightarrow \left| {x + 1} \right| = 2x + \dfrac{1}{2}\\
     \Leftrightarrow \left\{ \begin{array}{l}
    2x + \dfrac{1}{2} \ge 0\\
    \left[ \begin{array}{l}
    x + 1 = 2x + \dfrac{1}{2}\\
    x + 1 =  – 2x – \dfrac{1}{2}
    \end{array} \right.
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    x \ge  – \dfrac{1}{4}\\
    \left[ \begin{array}{l}
    x = \dfrac{1}{2}\\
    x =  – \dfrac{1}{2}
    \end{array} \right.
    \end{array} \right. \Leftrightarrow x = \dfrac{1}{2}\\
    c,\\
    \left| {2x – 1} \right| – \left| {x + \dfrac{1}{3}} \right| = 0\\
     \Leftrightarrow \left| {2x – 1} \right| = \left| {x + \dfrac{1}{3}} \right|\\
     \Leftrightarrow \left[ \begin{array}{l}
    2x – 1 = x + \dfrac{1}{3}\\
    2x – 1 =  – x – \dfrac{1}{3}
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = \dfrac{4}{3}\\
    x = \dfrac{2}{9}
    \end{array} \right.
    \end{array}\)

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