Bài 1:Tìm x biết:
2/1x3+2/3x5+2/5x7+.....+2/x(x+2)=2015/2016
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DD
Đoàn Đức Hà
Giáo viên
25 tháng 5 2021
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 5 2021
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
VH
0
DB
3
VN
Võ Ngọc Phương
VIP
8 tháng 8 2023
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
SL
0
NL
1
21 tháng 8 2019
\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).y=\frac{2}{3}\)
\(2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)
\(2.\frac{10}{11}.y=\frac{2}{3}\)
\(\frac{20}{11}.y=\frac{2}{3}\)
\(\Rightarrow y=\frac{11}{30}\)
Study well
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
NT
1
31 tháng 1 2022
\(\Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{101}{102}\)
\(\Leftrightarrow\dfrac{x+2-1}{x+2}=\dfrac{101}{102}\)
=>x+1=101
hay x=100
D
4
S
27 tháng 2 2017
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)
=> \(\frac{5}{11}y=\frac{2}{3}\)
=>y = \(\frac{2}{3}:\frac{5}{11}\)
=> y = \(\frac{22}{15}\)
3 tháng 4 2021
cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm
5 tháng 6 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}+\frac{1}{2015.2017}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}+\frac{2}{2015.2017}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2017}\)
\(2A=1-\frac{1}{2017}\)
\(2A=\frac{2016}{2017}\)
\(A=\frac{2016}{2017}:2\)
\(A=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{x}-\frac{2}{\left(x+2\right)}=\frac{2015}{2016}\)
\(\Rightarrow2-\frac{2}{x+2}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{x+2}=2-\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{x+2}=\frac{2017}{2016}\)
\(\Rightarrow2017.\left(x+2\right)=2.2016\)
\(\Rightarrow2017x+4034=4032\)
\(\Rightarrow2017x=-2\)
\(\Rightarrow x=-\frac{2}{2017}\)
Vậy......
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\cdot\left(x+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(=1-\frac{1}{x+2}=\frac{2015}{2016}\)
=>\(\frac{1}{x+2}=\frac{1}{2016}\)
=>\(x+2=2016\)
=>\(x=2014\)
Vậy.......