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= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\left(\frac{1}{2}\right)^2\right)^6\cdot2017}{\left(\frac{1}{2}\right)^2\cdot\frac{1}{3}\cdot\left(\frac{1}{2}\right)^{13}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot2018-\left(\frac{1}{2}\right)^{12}\cdot2017}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
=\(\frac{\left(\frac{1}{2}\right)^2\cdot\left(2018-2017\right)\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}.\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^2\cdot1\cdot\left(\frac{1}{2}\right)^{10}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{\left(\frac{1}{2}\right)^{12}}{\left(\frac{1}{2}\right)^{15}\cdot\frac{1}{3}}\)
= \(\frac{1}{\left(\frac{1}{2}\right)^3\cdot\frac{1}{3}}\)
= \(\frac{1}{\frac{1}{24}}\)
ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1
=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2
=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019
=> B= 2019 *(1/2+1/3+...+1/2019)
=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)
=> A/B= 1/2019
B= \(\frac{2\cdot2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}}\)
Ta có:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}\)
\(=1+\frac{1}{\frac{3.2}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{2018.2019}{2}}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2018.2019}\)
\(=\frac{2}{2}+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(=2\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)=\frac{2.2018}{2019}\)
=> B= \(\frac{2\cdot2018}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2018}}=\frac{2.2018}{\frac{2.2018}{2019}}=2019\)