x/1x2+x/2x3+x/3X4+........+x/2020x2021=-1
K
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18 tháng 9 2021
Nguồn: Tính tổng: 1x2 + 2x3 + 3x4 +...+ 2019x2020 + 2020x2021 - Hoc24
Đặt
KT
2
XO
18 tháng 10 2020
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2019.2020 + 2020.2021
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2019.2020.3 + 2020.2021.3
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2019.2020.(2021 - 2018) + 2020.2021.(2022 - 2019)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2019.2020.2021 - 2018.2019.2020 + 2020.2021.2022 - 2019.2020.2021
=> 3A = 2020.2021.2022
=> A = 2 751 551 080
NN
18 tháng 10 2020
Đặt \(A=1.2+2.3+3.4+.........+2019.2020+2020.2021\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+.....+2019.2020.3+2020.2021.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+2020.2021.\left(2022-2019\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2020.2021.2022-2019.2020.2021\)
\(=2020.2021.2022\)
\(\Rightarrow A=\frac{2020.2021.2022}{3}\)
NC
23 tháng 8 2020
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)
XO
19 tháng 8 2020
a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)
b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)
c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)
e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
BD
2
11 tháng 5 2021
S= 2x(1/1x2+1/2x3+1/3x4+...........+1/2020x2021)
S=2x(1-1/2+1/2-1/3+1/3-...+1/2020-1/2021)
S=2x(1-1/2021)
S=2x2020/2021
S=4040/2021
2019/2010<3/2<4040/2021
=>2019/2010<S
E
11 tháng 5 2021
S = 2 x (\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\)\(\frac{2}{2020\times2021}\))
= 2 x (\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\)\(\frac{1}{2020\times2021}\))
= 2 x ( \(1-\frac{1}{2021}\))
= \(2\times\frac{2020}{2021}\)
= \(\frac{4040}{2021}\)
= \(\frac{4042-2}{2021}\)
\(=2-\frac{2}{2021}\)
Ta có :
\(\frac{2019}{2010}=\frac{2020-1}{2010}=2-\frac{1}{2010}=2-\frac{2}{2020}\)
Ta thấy \(\frac{2}{2021}< \frac{2}{2020}\)
nên \(2-\frac{2}{2021}>2-\frac{2}{2020}\)
Vậy \(S\)\(>\frac{2019}{2010}\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2023
Đề đã đầy đủ chưa bạn? Và bạn đang cần làm gì với biểu thức này?
CB
2
HT
25 tháng 7 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
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) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}\right)=-1\)
\(\Leftrightarrow x\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2021-2020}{2020.2021}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2021}\right)=-1\Leftrightarrow x.\frac{2020}{2021}=-1\Rightarrow x=-\frac{2021}{2020}\)