A=(1-1/2)×(1-1/3)×(1-1/4)×(1-1/5)×(1-1/2016)×(1-1/2017)
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JC
3
2 tháng 8 2016
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{2}{3}\right)...\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(A=\frac{1.2.3...2016}{2.3.4...2017}=\frac{1}{2017}\)
2 tháng 8 2016
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2016}\right)\cdot\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2015}{2016}\cdot\frac{2016}{2017}\)
\(A=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2015\cdot2016}{2\cdot3\cdot4\cdot5\cdot....\cdot2016\cdot2017}\)
\(A=\frac{1}{2017}\)
NT
2
16 tháng 5 2018
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)
\(A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2017-2015}{2015.2016.2017}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2A=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\div2\)
DN
0
PA
0
PT
1
21 tháng 6 2016
\(A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2015}{2016}\times\frac{2016}{2017}=\frac{1}{2017}\)
PL
1
9 tháng 5 2019
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)