bài 1:tìm X
phần a, 1/1x2 + 1/ 2x 3 + 1/ 3 x4 + ....+ 1/ X x( X +1)= 499 / 500
phần b,1/1x3 + 1/3x5 + 1/ 5 x 7+ ....+ 1/X x ( X + 2 )= 20 /41
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Những câu hỏi liên quan
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) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
16 tháng 9 2018
M = 5 + 53 + 55 + ... + 547 + 549
52M = 52(5 + 53 + 55 + ... + 547 + 549)
25M = 53 + 55 + 57 + ... + 549 + 551
25M - M = ( 53 + 55 + 57 + ... + 549 + 551) - (5 + 53 + 55 + ... + 547 + 549)
24M = 551 - 5
M = \(\frac{5^{51}-5}{24}\)
9 tháng 8 2018
a) A = 2 + 4 + 6 + 8 + ... + 1000
Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )
= (1000 + 2) . 500 : 2 = 250500
c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
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) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)
O
3 tháng 5 2022
a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)
\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)
\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)
\(=\dfrac{2}{9}\)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{9}{9}-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
O
3 tháng 5 2022
Sửa câu b)
b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
Đặt \(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)
\(2A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)
\(2A=1-\dfrac{1}{9}\)
\(2A=\dfrac{9}{9}-\dfrac{1}{9}\)
\(2A=\dfrac{8}{9}\)
\(A=\dfrac{8}{9}:2\)
\(A=\dfrac{8}{18}\)
\(A=\dfrac{4}{9}\)
Vậy : \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}=\dfrac{4}{9}\)
DB
3
VN
Võ Ngọc Phương
CTVHS
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: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
7 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+3\right)}=\frac{20}{41}\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{20}{41}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 41 - 2
=> x = 39
Vẫy x = 39
HT
7 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
PT
3
C
29 tháng 5 2019
\(\left(x\cdot2,4-4,2\right)\div x=1\)
\(\Rightarrow x\cdot2,4-4,2=x\)
\(\Rightarrow x\cdot2,4=x+4,2\)
\(\Rightarrow\frac{12x}{5}=\frac{5x+21}{5}\)
\(\Rightarrow12x=5x+21\)
\(\Rightarrow12x-5x=21\)
\(\Rightarrow7x=21\Rightarrow x=\frac{21}{7}=3\)
Vậy x = 3
C
29 tháng 5 2019
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{x+1}{2x+2}\)
16 tháng 8 2019
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
16 tháng 8 2019
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
a) 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
1/x+1 = 1 - 499/500
1/x+1 = 1/500
x + 1 = 500
x = 500 - 1
x = 499
b) 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/x.(x+2) = 20/41
1/2 . [ 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2) ] = 20/41
1/2 . [ 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 ] = 20/41
1/2 . [ 1 - 1/x+2 ) = 20/41
1 - 1/x+2 = 20/41 : 1/2
1 - 1/x+2 = 40/41
1/x+2 = 1 - 40/41
1/x+2 = 1/41
x + 2 = 41
x = 41 - 2
x = 39