\(10+\frac{1}{12}\times100\)
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AS
4
ML
23 tháng 6 2017
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
DG
19 tháng 4 2017
\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)
19 tháng 4 2017
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)
\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)
\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)
\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)
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Đúng 100%
12 tháng 3 2017
A = \(\left(\frac{1}{11}+\frac{1}{12}+.........+\frac{1}{20}\right)\) + \(\left(\frac{1}{21}+\frac{1}{22}+..........+\frac{1}{30}\right)\)+ \(\left(\frac{1}{31}+.....+\frac{1}{60}\right)\)+ ... + \(\frac{1}{70}\)
Nhận xét:
\(\frac{1}{11}\)+ \(\frac{1}{12}\)+ ........ + \(\frac{1}{20}\)> \(\frac{1}{20}\)+\(\frac{1}{20}\)+........+\(\frac{1}{20}\)> \(\frac{10}{20}\)>\(\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+.......+\frac{1}{30}>\frac{30}{60}>\frac{1}{2}\)
\(\frac{1}{31}+......+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+.......+\frac{1}{60}>\frac{30}{60}>\frac{1}{2}\)
A > \(\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+......+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}>\frac{4}{3}\)
DP
13 tháng 7 2017
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
TH
13 tháng 7 2017
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
KD
3
25 tháng 1 2017
\(A=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{99\times100}\)
\(A=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{10}-\frac{1}{100}\)
\(A=\frac{9}{100}\)
PT
25 tháng 1 2017
\(A=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{99.100}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{10}-\frac{1}{100}\)
\(=\frac{9}{100}\)
7 tháng 5 2018
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
7 tháng 5 2018
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
6 tháng 8 2016
\(\frac{1}{10}.1234:\frac{3}{12}\)
\(=\frac{617}{5}:\frac{3}{12}\)
\(=\frac{2468}{5}\)
10 + 1/12 × 100
= 10 + 25/3
= 30/3 + 25/3
= 55/3.
\(10+\frac{1}{12}\cdot100=100+\frac{25}{3}\)
\(=\frac{300}{3}+\frac{25}{3}\)
\(=\frac{325}{3}\)