1/1.4+1/4.7+1/7.10+.......+1/16.19
1/10=1/15+1/21+........+1/120
Tính nhanh (nếu có thể).
K
Khách
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M
2
Những câu hỏi liên quan
2 tháng 10 2023
`#3107.101107`
1.
a)
`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`
`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`
`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`
`= 1/3* (1 - 1/103)`
`= 1/3*102/103`
`= 34/103`
b)
`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`
`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`
`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`
`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`
`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`
`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`
`= -1/2 * (1 - 1/101)`
`= -1/2*100/101`
`= -50/101`
2.
`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`
`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`
`= 1-1/100`
`= 99/100`
NM
Nguyễn Minh Quang
Giáo viên
10 tháng 3 2022
ta nhân 3 cả hai vế, được :
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)
hay
\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)
\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)
ND
Nguyễn Đức Trí
VIP
20 tháng 8 2023
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2021.2014}\)
\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2014}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{2014}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\dfrac{2013}{2014}=\dfrac{671}{2014}\)
HQ
Hà Quang Minh
Giáo viên
20 tháng 8 2023
\(B=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{2021\cdot2024}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2021\cdot2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\dfrac{2023}{2024}\\ =\dfrac{2023}{6072}\)
20 tháng 8 2023
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
DV
1
B
3
G
4 tháng 2 2020
A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)
YN
4 tháng 2 2020
Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)
Vậy \(A=\frac{672}{2017}\)
~ Học tốt
# Chiyuki Fujito
NT
2
28 tháng 1 2022
\(B=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2008}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{3}.\dfrac{2010}{2011}=\dfrac{2010}{6033}\)
Lại có : \(1=\dfrac{6033}{6033}\Rightarrow B< 1\)
28 tháng 1 2022
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2008.2011}\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2008}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{3}.\dfrac{2010}{2011}\)
\(=\dfrac{2010}{6033}=\dfrac{670}{2011}\)
Vì phân số \(\dfrac{670}{2011}\) có tử số nhỏ hơn mẫu số ⇒ \(\dfrac{670}{2011}< 1\) hay \(B< 1\)
26 tháng 12 2015
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}=\frac{0,33\cdot x}{2009}\cdot3\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,99\cdot x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,99x}{2009}\)
\(\frac{99}{100}=\frac{0,99x}{2009}\)
=>0,99x*100=2009*99
99x=2009*99
=>x=2009
Vậy x=2009
26 tháng 12 2015
\(0,33\cdot\frac{x}{2009}\) hay \(\frac{0,33\cdot x}{2009}\)
1/1.4+1/4.7+1/7.10+...+1/16.19
=[1/1.4+1/4.7+1/7.10+...+1/16.19] x 3
= 3/1.4+3/4.7+3/7.10+...+3/16.19
= 1-1/4+1/4-1/7+1/7-1/10+....+1/16-1/19
=1-1/19
=18/19 :3
=6/19
ĐÂY BẠN NHÉ CHÚC BẠN HỌC TỐT NHỚ K CHO MÌNH
A = \(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\)
A= \(\dfrac{2}{2}.\left(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\right)\)
A= \(2.\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{240}\right)\)
A= \(2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)
A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
A=\(\dfrac{2.3}{16}\)
A= \(\dfrac{3}{8}\)