[tex]\text{Pnetru inceput sa rezolvam suma din paranteza.O voi nota cu S.}\\
S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{2^{2017}}|\cdot \dfrac{1}{2}\\
\dfrac{1}{2}S=\dfrac{1}{2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{2^{2018}}\\
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S-\dfrac{1}{2}S=1-\dfrac{1}{2^{2018}}\\
\dfrac{1}{2}S=1-\dfrac{1}{2^{2018}}\\
S=\dfrac{2^{2018}-1}{2^{2017}}}[/tex]
[tex]\text{Revenind la exercitiu:}\\
x:\dfrac{2^{2018}-1}{2^{2017}}=\dfrac{2^{2016}}{2^{2018}-1}\\
x\cdot \dfrac{2^{2017}}{2^{2018}-1}=\dfrac{2^{2016}}{2^{2018}-1}\\
x\cdot 2^{2017}=2^{2016}\\
\boxed{x=\dfrac{1}{2}}[/tex]
