tìm x 1/1x2+1/23+1/3x4.....1/Xx(X+1)=1/2
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14 tháng 8 2017
Ta có : A = \(\frac{1}{1\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{3\text{x}4}+...+\frac{1}{X\text{x}\left(X+1\right)}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
A = \(\frac{1}{1}-\frac{1}{x+1}\)
A = \(\frac{x}{x+1}\)
Ủng hộ mik nhá !!!!
TP
14 tháng 8 2017
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-?\)
\(\Rightarrow x+1=?\Leftrightarrow x=?\)
NQ
1
13 tháng 6 2015
ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
bài toán được viết lại như sau:
\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)
\(\Rightarrow x=2012\)
vậy x=2012
7 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{499}{500}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{499}{500}\)
\(1-\frac{1}{x+1}=\frac{499}{500}\)
\(\frac{1}{x+1}=1-\frac{499}{500}=\frac{1}{500}\)
=> x + 1 = 500
=> x = 500 - 1
=> x = 499
Vậy x = 499
7 tháng 8 2016
1/1.2 + 1/2.3 + 1/3.4 +...+ 1/x.(x+1)=499/500
1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 +...+ 1/x -1/(x+1) =499/500
1-1/(x+1)=499/500
=>x/(x+1)=499/500
=>x=499
7 tháng 8 2016
Ta có: 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/X x (X + 1) = 499/500
=> 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/X - 1/(X + 1) = 499/500
=> 1 - 1/(X + 1) = 499/500
=> 1/(X + 1) = 1 - 499/500
=> 1/(X + 1) = 1/500
=> X + 1 = 500
=> X = 500 - 1
=> X = 499
Đáp số: X = 499
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
DK
2
VN
Võ Ngọc Phương
VIP
24 tháng 8 2023
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{\left(x-1\right)\times x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{15}{16}\)
\(1-\dfrac{1}{x}=\dfrac{15}{16}\)
\(\dfrac{1}{x}=1-\dfrac{15}{16}=\dfrac{16}{16}-\dfrac{15}{16}\)
\(\dfrac{1}{x}=\dfrac{1}{16}\)
\(\Rightarrow x=16\)
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) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
NT
2
WH
11 tháng 3 2018
Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)
\(\Rightarrow x+1=2014\)
\(\Rightarrow x=2014-1\)
\(\Rightarrow x=2013\)
Vậy x=2013
T
11 tháng 3 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\frac{1}{x+1}=\frac{1}{2014}\)
Vì \(x+1\)là mẫu số nên:
\(x+1=2014\)
\(x=2014-1=2013\)
Vậy ....
P/s: Dấu . là nhân nha!
NN
3 tháng 7 2015
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{2}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{1}{2}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow x=1\)