TÌM X:2^x+2^x+1+2^x+2+2^x+3+…+2^x+2015=2^2019-8

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1. Đáp án: x=3

    

   Giải thích các bước giải:

   $\begin{array}{l}
   A = {2^x} + {2^{x + 1}} + {2^{x + 2}} + … + {2^{x + 2015}}\\
    \Rightarrow 2A = {2^{x + 1}} + {2^{x + 1}} + {2^{x + 3}} + .. + {2^{x + 2016}}\\
    \Rightarrow 2A – A = {2^{x + 2016}} – {2^x}\\
    \Rightarrow A = {2^x}{.2^{2016}} – {2^x}\\
    \Rightarrow {2^x}{.2^{2016}} – {2^x} = {2^{2019}} – 8 = {2^{2019}} – {2^3}\\
    \Rightarrow x = 3
   \end{array}$

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2. Em tham khảo

   $N=2^x+2^{x+1}+2^{x+2}+….+2^{x+2015}$

   $⇒2N=2^{x+1}+2^{x+2}+….+2^{x+2016}$

   $⇒2N-N=(2^{x+1}+2^{x+2}+….+2^{x+2016})-(2^x+2^{x+1}+2^{x+2}+….+2^{x+2015})$

   $⇒N=2^{x+2016}-2^x$

   $⇒2^{x+2016}-2x=2^{2019}-8$

   hay $2^{x+2016}-2x=2^{2019}-2^3$

   $⇒x=3$

   $\text{Xin hay nhất}$

    

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