K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
0
VN
4 tháng 2 2018
Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
4 tháng 2 2018
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
NT
1
HN
15 tháng 7 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 1 2024
(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)
(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1 - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))
\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = \(\dfrac{1}{2003}\)
\(x\) = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)
\(x\) = \(\dfrac{2022}{2003}\)
TH
1
31 tháng 1 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(x-1\right)x}=2\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x-1}-\frac{1}{x}=2\)
suy ra \(1-\frac{1}{x}=2\)
hay \(\frac{x-1}{x}=2\) .suy ra x-1=2x .tính ra ta có x=-1
22 tháng 11 2015
a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n
3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]
3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n
3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)
A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)
b) Đặt B = 12 + 22 + ..... + n2
B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]
B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)
B = A - \(\frac{n\left(n+1\right)}{2}\)
NC
2
S
16 tháng 4 2017
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
BT
16 tháng 4 2017
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
NT
1
1 tháng 8 2019
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{45.46}=-2\)
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{45.46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{45}-\frac{1}{46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{46}\right)\)
\(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=-2+\frac{45}{46}\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(2x=\frac{46}{-47}\)
\(x=\frac{46}{-47}:2=\frac{-23}{47}\)
NT
0
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-......-\frac{1}{45.46}=-2\)2
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{45.46}\right)=-2\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{45.46}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{45}-\frac{1}{46}\)
\(A=1-\frac{1}{46}=\frac{45}{46}\)
Ta có: \(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(\frac{-47}{-94.x}=\frac{-47}{46}\)
\(\Rightarrow x=\frac{-23}{47}\)