bài 1:tìm x a ($\frac{4}{3}$ -$\frac{1}{2}$.x )=$\frac{49}{9}$ b $x^{3}$ =$x^{2}$ c $27^{x}$ :$3^{x}$ =9 d $64^{x}$ :$4^{x^+2}$= 16 e $\frac{81}

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1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   \dfrac{4}{3} – \dfrac{1}{2}x = \dfrac{{49}}{9}\\
    \Leftrightarrow \dfrac{1}{2}x = \dfrac{4}{3} – \dfrac{{49}}{9}\\
    \Leftrightarrow \dfrac{1}{2}x =  – \dfrac{{37}}{9}\\
    \Leftrightarrow x = \left( { – \dfrac{{37}}{9}} \right):\dfrac{1}{2}\\
    \Leftrightarrow x =  – \dfrac{{74}}{9}\\
   b,\\
   {x^3} = {x^2}\\
    \Leftrightarrow {x^3} – {x^2} = 0\\
    \Leftrightarrow {x^2}\left( {x – 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   {x^2} = 0\\
   x – 1 = 0
   \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
   x = 0\\
   x = 1
   \end{array} \right.\\
   c,\\
   {27^x}:{3^x} = 9\\
    \Leftrightarrow {\left( {27:3} \right)^x} = 9\\
    \Leftrightarrow {9^x} = 9\\
    \Leftrightarrow x = 1\\
   d,\\
   {64^x}:{4^{x + 2}} = 16\\
    \Leftrightarrow {\left( {{4^3}} \right)^x}:{4^{x + 2}} = {4^2}\\
    \Leftrightarrow {4^{3x}}:{4^{x + 2}} = {4^2}\\
    \Leftrightarrow {4^{3x – \left( {x + 2} \right)}} = {4^2}\\
    \Leftrightarrow {4^{2x – 2}} = {4^2}\\
    \Leftrightarrow 2x – 2 = 2\\
    \Leftrightarrow x = 2\\
   e,\\
   \dfrac{{81}}{{{3^x}}} = 9\\
    \Leftrightarrow {3^x} = 81:9\\
    \Leftrightarrow {3^x} = 9\\
    \Leftrightarrow {3^x} = {3^2}\\
    \Leftrightarrow x = 2
   \end{array}\)

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