Giải:
\(\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left(\dfrac{1}{4}\right)^6.2017}{\dfrac{1}{4096}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left[\left(\dfrac{1}{2}\right)^2\right]^6.2017}{\left(\dfrac{1}{2}\right)^{12}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.2018-\left(\dfrac{1}{2}\right)^{12}.2017}{\left(\dfrac{1}{2}\right)^{12}.\dfrac{1}{3}-\left(\dfrac{1}{2}\right)^{13}}\)
\(=\dfrac{\left(\dfrac{1}{2}\right)^2.\left[2018-\left(\dfrac{1}{2}\right)^{10}.2017\right]}{\left(\dfrac{1}{2}\right)^{12}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)}\)
\(=\dfrac{2018-\left(\dfrac{1}{2}\right)^{10}.2017}{\left(\dfrac{1}{2}\right)^{10}.\left(-\dfrac{1}{6}\right)}\)
\(=\dfrac{2018}{\left(\dfrac{1}{2}^{10}\right).\left(-\dfrac{1}{6}\right)}-\dfrac{\left(\dfrac{1}{2}\right)^{10}.2017}{\left(\dfrac{1}{2}\right)^{10}.\left(-\dfrac{1}{6}\right)}\)
\(=\dfrac{2018}{\left(\dfrac{1}{2}^{10}\right).\left(-\dfrac{1}{6}\right)}+\dfrac{2017}{\dfrac{1}{6}}\)
\(=-12398592+12102\)
\(=-12386490\)
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