NH
3
K
Khách
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DT
2
N
25 tháng 3 2017
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\frac{100}{101}\)
\(=\frac{350}{101}\)
k mk nha
25 tháng 3 2017
7/1.3+7/3.5+7/5.7+...+7/99.101
=7(1/1.3+1/3.5+1/5.7+...+1/99.101)
=7(1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=7(1-1/101)
=7.100/101
=700/101
Đầy đủ ko bỏ bước nào lun!!
K CHO MK NHA!!!
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 5 2022
Đặt tông trên là A
\(\dfrac{2A}{7}=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
\(\Rightarrow A=\dfrac{7.2022}{2.2023}=\dfrac{1011}{289}\)
1 tháng 7 2015
\(\Leftrightarrow2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)+4x=7.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)\(\Leftrightarrow2.\left(1-\frac{1}{49}\right)+4x=7.\left(1-\frac{1}{99}\right)\)
\(\Leftrightarrow2.\frac{48}{49}+4x=7.\frac{98}{99}\)
\(\Leftrightarrow\frac{96}{49}+4x=\frac{686}{99}\)
\(\Leftrightarrow4x=\frac{686}{99}-\frac{96}{49}\)
\(\Leftrightarrow4x=4,970109256\)
\(\Leftrightarrow x=4,970109256:4\)
\(\Leftrightarrow x=1,242527314\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
5 tháng 5 2022
Đặt tổng trên là A
\(\dfrac{2A}{7}=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}=\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=\)
\(=\dfrac{2022}{2023}\Rightarrow A=\dfrac{7.2022}{2.2023}\)
MH
5 tháng 5 2022
\(A=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{2021}-\dfrac{2}{2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{2023}=\dfrac{4044}{2023}\Rightarrow A=\dfrac{2022}{289}\)
NT
4
TY
3 tháng 3 2017
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\).
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)
\(2A=\frac{1}{1}-\frac{1}{201}\)
\(2A=\frac{201-1}{201}\)
\(2A=\frac{200}{201}\)
\(A=\frac{200}{201}:2\)
\(A=\frac{200}{402}\)
Q
6 tháng 3 2017
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-...-\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\)
\(=7.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{7}{5.7}+...+\frac{7}{99.101}\right)\)
\(=7.\frac{1}{7}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{7}\left(1-\frac{1}{101}\right)\)
\(=\frac{100}{101}\)
NH
1
PD
11 tháng 3 2017
ta có:
201.3 + 2/3.5+2/5.7+..................+2/199.201
= 201.3+ 2/2 . ( 1/3 -1/5)+2/2.(1/5-1/7)+2/2.(1/5-1/7) +..............+2/2.(1/199.1/201)
201.3 +2/2.1/3.-2/2.1/5+2/2.1/5-2/2.1/7..........................+2/2.1/199-2/2.1/201
=201.3 +2/2.(1/3+1/5-1/5+1/7-1/7.................+1/99.1/201)
=201.3+2/2.(1/3-1/201)
=201.3+22/67
=198
^-^
7 tháng 3 2017
\(\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+....+\frac{7}{99.101}\)
\(=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)
PT
4
NH
17 tháng 6 2019
\(D=\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}.\frac{200}{201}\)
\(D=\frac{100}{67}\)
T
17 tháng 6 2019
#)Giải :
\(D=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{199.201}\)
\(D=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\left(1-\frac{1}{201}\right)\)
\(D=\frac{3}{2}\times\frac{200}{201}\)
\(D=\frac{100}{67}\)
G=7/2.(2/1.3+2/3.5+2/5.7+...+2/199.201)
G=7/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/199.201
G=7/2.(1-1/201)
G=7/2.200/201
G=1400/402
G=700/201
\(G=\frac{7}{1.3}+\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{199.201}\)
\(G=7\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.200}\right)\)
\(G=\frac{7}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{199.201}\right)\)
\(G=\frac{7}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(G=\frac{7}{2}\left(1-\frac{1}{201}\right)\)
\(G=\frac{7}{2}.\frac{200}{201}\)
\(G=\frac{700}{201}\)