Bài 2: C1:2.3^x=19.3^8-81^2 C2:2^x+2-2^x=48 C3:5^x=5^2019:(5^2013-100.5^2010) C4:(5^2+3^2).x+(5^2-3^2).x-40.x=10^2

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1. Giải thích các bước giải:

   \[\begin{array}{l}
   1,\\
   {2.3^x} = {19.3^8} – {81^2}\\
    \Leftrightarrow {2.3^x} = {19.3^8} – {\left( {{3^4}} \right)^2}\\
    \Leftrightarrow {2.3^x} = {19.3^8} – {3^8}\\
    \Leftrightarrow {2.3^x} = {18.3^8}\\
    \Leftrightarrow {3^x} = {9.3^8}\\
    \Leftrightarrow {3^x} = {3^{10}}\\
    \Leftrightarrow x = 10\\
   2,\\
   {2^{x + 2}} – {2^x} = 48\\
    \Leftrightarrow {2^x}{.2^2} – {2^x} = {3.2^4}\\
    \Leftrightarrow {3.2^x} = {3.2^4}\\
    \Leftrightarrow x = 4\\
   3,\\
   {5^x} = {5^{2019}}:\left( {{5^{2013}} – {{100.5}^{2010}}} \right)\\
    \Leftrightarrow {5^x} = {5^{2019}}:\left( {{5^{2013}} – {{4.5}^2}{{.5}^{2010}}} \right)\\
    \Leftrightarrow {5^x} = {5^{2019}}:\left( {{{5.5}^{2012}} – {{4.5}^{2012}}} \right)\\
    \Leftrightarrow {5^x} = {5^{2019}}:{5^{2012}}\\
    \Leftrightarrow {5^x} = {5^7}\\
    \Leftrightarrow x = 7\\
   4,\\
   \left( {{5^2} + {3^2}} \right)x + \left( {{5^2} – {3^2}} \right)x – 40x = {10^2}\\
    \Leftrightarrow x\left( {{5^2} + {3^2} + {5^2} – {3^2} – 40} \right) = 100\\
    \Leftrightarrow x.\left( {{{2.5}^2} – 40} \right) = 100\\
    \Leftrightarrow x.10 = 100\\
    \Leftrightarrow x = 10
   \end{array}\]

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