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\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)
\(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)
\(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)
\(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)
Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)