Tính tổng : 3/1.4+3/4.7+3/7.10+.......+3/2008.2011
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28 tháng 3 2017
a,1/1-1/4+1/4-1/7+...+1/2008-1/2011
=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)
=1-1/2011+0+...+0
=1-1/2011
=2010/2011
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 \(B< 1\)
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`
17 tháng 6 2020
Sửa đề : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
NM
3
MT
5 tháng 8 2015
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
6 tháng 3 2018
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
11 tháng 2 2018
Ta có :
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=\)\(1-\frac{1}{43}\)
\(=\)\(\frac{42}{43}\)
20 tháng 9 2016
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(\Rightarrow A=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(\Rightarrow A=3\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(\Rightarrow A=3\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=3.\frac{99}{100}\)
\(\Rightarrow A=\frac{297}{100}\)
DQ
1
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{2008.2011}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2008}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}\)
\(=\frac{2011}{2011}-\frac{1}{2011}\)
\(=\frac{2010}{2011}\)
Chúc bạn học tốt !!!!
Đặt: A= \(\frac{3}{1\times4}\)+ \(\frac{3}{4\times7}\)+ \(\frac{3}{7\times10}\)+...+ \(\frac{3}{2005\times2008}\)+ \(\frac{3}{2008\times2011}\).
A= \(\frac{3}{1}\)- \(\frac{3}{4}\)+ \(\frac{3}{4}\)- \(\frac{3}{7}\)+ \(\frac{3}{7}\)- \(\frac{3}{10}\)+...+ \(\frac{3}{2005}\)- \(\frac{3}{2008}\)+ \(\frac{3}{2008}\)- \(\frac{3}{2011}\).
A= 3- \(\frac{3}{2011}\).
A= \(\frac{6033}{2011}\)- \(\frac{3}{2011}\).
A= \(\frac{6030}{2011}\).
Vậy A= \(\frac{6030}{2011}\).