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3 tháng 2 2020
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004.2005}\)
Ta có: \(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004.2005}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{2004.2005}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2004.2005}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(A=\frac{2003}{2005}\)
20 tháng 5 2016
Đặt A = 1/2 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/36 + 1/45
=> 1/2A = 1/4 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/72 + 1/90
= 1/4 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/8.9 + 1/9.10
= 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
= 1/4 + 1/2 - 1/10
= 5/20 + 10/20 - 2/20
= 13/20
=> A = 13/20 : 1/2 = 13/10
QD
20 tháng 5 2016
Đặt A = 1/2 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/36 + 1/45
=> 1/2A = 1/4 + 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/72 + 1/90
= 1/4 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/8.9 + 1/9.10
= 1/4 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
= 1/4 + 1/2 - 1/10
= 5/20 + 10/20 - 2/20 = 13/20
=> A = 13/20 : 1/2 = 13/10
DL
13 tháng 5 2019
Ta co:
\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{4}+\frac{1}{2}-\frac{1}{10}=\frac{13}{20}\Rightarrow A=\frac{13}{10}.\)
13 tháng 5 2019
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{36}+\frac{1}{45}\)
\(A=\frac{2}{4}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{72}+\frac{2}{90}\)
\(A=\frac{2}{2.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{8.9}+\frac{2}{9.10}\)
\(A=2\left(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=2.\frac{2}{5}\)
\(A=\frac{4}{5}\)
~ Học tốt ~ K cho mk nhé! Thank you.
31 tháng 1 2022
a: \(=\left(\dfrac{2}{18}-\dfrac{15}{18}-\dfrac{72}{18}\right):\left(\dfrac{21}{36}-\dfrac{1}{36}-\dfrac{360}{36}\right)\)
\(=\dfrac{-85}{18}:\dfrac{-170}{18}\)
\(=\dfrac{85}{170}=\dfrac{1}{2}\)
b: \(=\left(\dfrac{5}{8}-\dfrac{5}{6}-\dfrac{5}{32}+\dfrac{5}{64}\right):\left(1-\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}\right)\)
\(=\dfrac{-55}{192}:\dfrac{3}{8}=\dfrac{-55}{192}\cdot\dfrac{8}{3}=-\dfrac{55}{72}\)
PQ
1
13 tháng 7 2016
A = -1 - 1/3 - 1/6 - 1/10 - 1/15 - ... - 1/1225
A = -(1 + 1/3 + 1/6 + 1/10 + 1/15 + ... + 1/1225)
A = -(2/2 + 2/6 + 2/12 + 2/20 + 2/30 + ... + 2/2450)
A = -2.(1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/49.50)
A = -2.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/49 - 1/50)
A = -2.(1 - 1/50)
A = -2.49/50
A = -49/25
L
8 tháng 3 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2004.2005}\)
\(A=\frac{1}{1.2}=1-\frac{1}{2}\)
\(A=\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(A=1-\frac{1}{2004}\)
\(A=\frac{2003}{2004}\)
Ủng hộ tk Đúng nha mọi người !!! ^^
BT
8 tháng 3 2017
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\);...; \(\frac{1}{2004.2005}=\frac{1}{2004}-\frac{1}{2005}\)
=> A=\(\frac{1}{1}-\frac{1}{2005}=\frac{2004}{2005}\)
10 tháng 8 2016
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{2004.2005}\)
\(\Leftrightarrow2M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{2004.2005}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{2004.2005}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2004}-\frac{1}{2005}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(=2.\left(\frac{2005}{4010}-\frac{2}{4010}\right)\)
\(=2.\frac{2003}{4010}\)
\(=\frac{2003}{2005}\)
\(M=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{2004\cdot2005}\)
\(M=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{2004\cdot2005}\)
\(M=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2004\cdot2005}\right)\)
\(M=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2004\cdot2005}\right)\)
\(M=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}\right)\)
\(M=2\left(\frac{1}{2}-\frac{1}{2005}\right)\)
\(M=2\cdot\frac{2003}{4010}\)
\(M=\frac{2003}{2005}\)