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LT
3
13 tháng 5 2016
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)
\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)
\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
....
\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
13 tháng 5 2016
\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)
=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)
=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)
=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
=\(\frac{5}{2}\cdot\frac{100}{101}\)
\(=\frac{250}{101}\)
2 tháng 3 2017
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
5 tháng 5 2019
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
29 tháng 4 2017
\(A=\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)
\(A=5.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)
\(A=5.\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)
\(A=\dfrac{5}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A=\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)
\(A=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{5.50}{101}=\dfrac{250}{101}=2\dfrac{48}{101}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
21 tháng 4 2016
Đặt BT trên là A
\(\frac{2}{5}.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{101-99}{99.101}\)
\(\frac{2}{5}.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\frac{100}{101}.\frac{5}{2}=\frac{250}{101}\)
25 tháng 3 2018
cái này bạn mở sách bồi dưỡng toán ra trang gần cuối là thấy ngay ấy mà
PT
4
P
11 tháng 8 2017
\(\frac{2}{5}.A\)= \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ ... + \(\frac{2}{99.100}\)= \(\frac{3-1}{1.3}\)+\(\frac{5-3}{3.5}\)+\(\frac{7-5}{5.7}\)+ ... + \(\frac{101-99}{99.101}\)
\(\frac{2}{5}.A\)= 1 \(-\)\(\frac{1}{3}\)+ \(\frac{1}{3}\)\(-\)\(\frac{1}{5}\)+\(\frac{1}{5}\)\(-\)\(\frac{1}{7}\)+ ... + \(\frac{1}{99}\)+\(\frac{1}{101}\)= 1\(-\)\(\frac{1}{101}\)=\(\frac{100}{101}\)
\(A\)=\(\frac{100}{101}\): \(\frac{2}{5}\)= \(\frac{100}{101}\).\(\frac{5}{2}\)= \(\frac{250}{101}\)
TV
11 tháng 8 2017
=5/2.(2/1.3+2/3.5+2/5.7+...+2/99.101)
=5/2.(1-1/3+1/3-1/5+1/5-1/7+..+1/99-1/101)
=5/2.(1-1/101)
=5/2.100/101
=250/101
S
1
17 tháng 4 2019
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
\(B=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(B=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(B=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(B=\frac{5}{2}.\frac{100}{101}\)
\(B=\frac{250}{101}\)
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(A=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{99.101}\right)\)
\(A=5.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{99.101}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{3}+1-\frac{3}{5}+1-\frac{5}{7}+1-\frac{99}{101}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{5}{2}.\frac{100}{101}=\frac{5.50}{101}=\frac{250}{101}2\frac{48}{101}\)
#Hokrot#
@. C.Ronaldo@
Em Sai từ dòng thứ 4 rồi nhé!
A=\(\frac{5}{2}\left(\frac{3-1}{3.1}+\frac{5-3}{3.5}+\frac{7-5}{7.5}+...+\frac{101-99}{101.99}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{101}\right)=\frac{250}{101}\)