f) 2(x + 1)/ 3 – 1 = 3/2 – 1-2x / 4 g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5 h) x^2 – x – 2 = 0

0 bình luận về “f) 2(x + 1)/ 3 – 1 = 3/2 – 1-2x / 4 g) 3(x + 1) / 10 = 2(x – 1) / 3 + 4/5 h) x^2 – x – 2 = 0”

1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   f,\\
   \dfrac{{2\left( {x + 1} \right)}}{3} – 1 = \dfrac{3}{2} – \dfrac{{1 – 2x}}{4}\\
    \Leftrightarrow \dfrac{2}{3}x + \dfrac{2}{3} – 1 = \dfrac{3}{2} – \left( {\dfrac{1}{4} – \dfrac{1}{2}x} \right)\\
    \Leftrightarrow \dfrac{2}{3}x – \dfrac{1}{3} = \dfrac{3}{2} – \dfrac{1}{4} + \dfrac{1}{2}x\\
    \Leftrightarrow \dfrac{2}{3}x – \dfrac{1}{2}x = \dfrac{3}{2} – \dfrac{1}{4} + \dfrac{1}{3}\\
    \Leftrightarrow \dfrac{1}{6}x = \dfrac{{19}}{{12}}\\
    \Leftrightarrow x = \dfrac{{19}}{2}\\
   g,\\
   \dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{2.\left( {x – 1} \right)}}{3} + \dfrac{4}{5}\\
    \Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{10.\left( {x – 1} \right) + 12}}{{15}}\\
    \Leftrightarrow \dfrac{{3\left( {x + 1} \right)}}{{10}} = \dfrac{{10x + 2}}{{15}}\\
    \Leftrightarrow 3.\left( {x + 1} \right).15 = 10.\left( {10x + 2} \right)\\
    \Leftrightarrow 3.\left( {x + 1} \right).3 = 2.\left( {10x + 2} \right)\\
    \Leftrightarrow 9\left( {x + 1} \right) = 2.\left( {10x + 2} \right)\\
    \Leftrightarrow 9x + 9 = 20x + 4\\
    \Leftrightarrow 9 – 4 = 20x – 9x\\
    \Leftrightarrow 5 = 11x\\
    \Leftrightarrow x = \dfrac{5}{{11}}\\
   h,\\
   {x^2} – x – 2 = 0\\
    \Leftrightarrow \left( {{x^2} – 2x} \right) + \left( {x – 2} \right) = 0\\
    \Leftrightarrow x\left( {x – 2} \right) + \left( {x – 2} \right) = 0\\
    \Leftrightarrow \left( {x – 2} \right)\left( {x + 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   x – 2 = 0\\
   x + 1 = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = 2\\
   x =  – 1
   \end{array} \right.
   \end{array}\)

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