mng giúp em vs …………………………………………….

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mng giúp em vs …………………………………………….
mng-giup-em-vs

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Vodka 4 years 2020-10-29T14:50:55+00:00 1 Answers 72 views 0

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  1. Acacia
    0
    2020-10-29T14:52:49+00:00
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    Giải thích các bước giải:

    Ta có:

    \(\begin{array}{l}
    a,\\
    4x\left( {x + 1} \right) = 8.\left( {x + 1} \right)\\
     \Leftrightarrow 4x.\left( {x + 1} \right) – 8.\left( {x + 1} \right) = 0\\
     \Leftrightarrow \left( {x + 1} \right).\left( {4x – 8} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x + 1 = 0\\
    4x – 8 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x =  – 1\\
    x = 2
    \end{array} \right.\\
    b,\\
    x\left( {x – 1} \right) – 2\left( {1 – x} \right) = 0\\
     \Leftrightarrow x\left( {x – 1} \right) + 2.\left( {x – 1} \right) = 0\\
     \Leftrightarrow \left( {x – 1} \right)\left( {x + 2} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 1 = 0\\
    x + 2 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 1\\
    x =  – 2
    \end{array} \right.\\
    c,\\
    2x\left( {x – 2} \right) – {\left( {2 – x} \right)^2} = 0\\
     \Leftrightarrow 2x\left( {x – 2} \right) – {\left( {x – 2} \right)^2} = 0\\
     \Leftrightarrow \left( {x – 2} \right).\left[ {2x – \left( {x – 2} \right)} \right] = 0\\
     \Leftrightarrow \left( {x – 2} \right)\left( {x + 2} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 2 = 0\\
    x + 2 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 2\\
    x =  – 2
    \end{array} \right.\\
    d,\\
    {\left( {x – 3} \right)^3} + 3 – x = 0\\
     \Leftrightarrow {\left( {x – 3} \right)^3} – \left( {x – 3} \right) = 0\\
     \Leftrightarrow \left( {x – 3} \right).\left[ {{{\left( {x – 3} \right)}^2} – 1} \right] = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 3 = 0\\
    {\left( {x – 3} \right)^2} – 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x – 3 = 0\\
    x – 3 = 1\\
    x – 3 =  – 1
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 3\\
    x = 4\\
    x = 2
    \end{array} \right.\\
    e,\\
    5x.\left( {x – 2} \right) – \left( {2 – x} \right) = 0\\
     \Leftrightarrow 5x.\left( {x – 2} \right) + \left( {x – 2} \right) = 0\\
     \Leftrightarrow \left( {x – 2} \right).\left( {5x + 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 2 = 0\\
    5x + 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 2\\
    x =  – \frac{1}{5}
    \end{array} \right.\\
    g,\\
    5x\left( {x – 2000} \right) – x + 2000 = 0\\
     \Leftrightarrow 5x\left( {x – 2000} \right) – \left( {x – 2000} \right) = 0\\
     \Leftrightarrow \left( {x – 2000} \right)\left( {5x – 1} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x – 2000 = 0\\
    5x – 1 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 2000\\
    x = \frac{1}{5}
    \end{array} \right.\\
    h,\\
    {x^2} – 4x = 0\\
     \Leftrightarrow x.\left( {x – 4} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    x – 4 = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    x = 4
    \end{array} \right.\\
    k,\\
    {\left( {1 – x} \right)^2} – 1 + x = 0\\
     \Leftrightarrow {\left( {1 – x} \right)^2} – \left( {1 – x} \right) = 0\\
     \Leftrightarrow \left( {1 – x} \right).\left[ {\left( {1 – x} \right) – 1} \right] = 0\\
     \Leftrightarrow \left( {1 – x} \right).\left( { – x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    1 – x = 0\\
     – x = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    x = 1
    \end{array} \right.\\
    m,\\
    x + 6{x^2} = 0\\
     \Leftrightarrow x.\left( {1 + 6x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    1 + 6x = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x = 0\\
    x =  – \frac{1}{6}
    \end{array} \right.\\
    n,\\
    \left( {x + 1} \right) = {\left( {x + 1} \right)^2}\\
     \Leftrightarrow \left( {x + 1} \right) – {\left( {x + 1} \right)^2} = 0\\
     \Leftrightarrow \left( {x + 1} \right).\left[ {1 – \left( {x + 1} \right)} \right] = 0\\
     \Leftrightarrow \left( {x + 1} \right).\left( { – x} \right) = 0\\
     \Leftrightarrow \left[ \begin{array}{l}
    x + 1 = 0\\
     – x = 0
    \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
    x =  – 1\\
    x = 0
    \end{array} \right.
    \end{array}\) 

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