TÍNH
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2015.2016}\)
K
Khách
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Những câu hỏi liên quan
10 tháng 5 2017
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
15 tháng 5 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
15 tháng 5 2017
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
TT
1
16 tháng 4 2016
S=2-1/1.2 . 3-2/2.3............2016-2015/2015.2016
=1/1 - 1/2 + 1/2 - 1/3+........+1/2015 - 1/2016
=1/1 - 1/2016
=2015/2016
7 tháng 5 2016
Ta có \(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...+\(\frac{1}{2015\cdot2016}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{2015}\)-\(\frac{1}{2016}\)
=1-\(\frac{1}{2016}\)
=\(\frac{2015}{2016}\)(bạn cứ nhớ công thức là làm được)
MC
7 tháng 5 2016
Ta thấy: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{2015.2016}=\frac{1}{2015}-\frac{1}{2016}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2015.2016}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
HB
0
BT
4 tháng 12 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)
\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)
=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)
=> \(B=\frac{1007}{4032}\)
MT
22 tháng 6 2015
\(\left(\frac{7}{1.2}+\frac{7}{2.3}+\frac{7}{3.4}+...+\frac{7}{2015.2016}\right):\frac{2015}{2016}\)
=\(7\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right):\frac{2015}{2016}\)
=\(7\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\frac{2015}{2016}\)
=\(7\left(\frac{1}{1}-\frac{1}{2016}\right):\frac{2015}{2016}=7.\frac{2015}{2016}:\frac{2015}{2016}=7\)
22 tháng 6 2015
\(\left(\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{2015\cdot2016}\right):\frac{2015}{2016}\)
\(=\left(7-\frac{7}{2}+\frac{7}{2}-\frac{7}{3}+\frac{7}{3}-\frac{7}{4}+...+\frac{7}{2015}-\frac{7}{2016}\right):\frac{2015}{2016}\)
\(=\left(7-\frac{7}{2016}\right):\frac{2015}{2016}=\frac{2015}{288}:\frac{2015}{2016}=\frac{2015}{288}\cdot\frac{2016}{2015}=\frac{2016}{288}=7\)
TT
2 tháng 5 2019
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
\(=\frac{2015}{2016}\)
2015/2016 nhé bạn.