B=1/1.4+1/4.7+1/7.10+......+1/2021.2014
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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`
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 \(\frac{0,33\cdot x}{2009}\)
2 tháng 5 2016
\(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)
\(\frac{1}{3}\times\frac{x+3-1}{x+3}=\frac{6}{19}\)
\(\frac{x+3-1}{x+3}=\frac{6}{19}\div\frac{1}{3}\)
\(\frac{x+2}{x+3}=\frac{18}{19}\)
x = 16
TM
8 tháng 7 2016
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
=>\(3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)
=>\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)
=>\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)
=>\(1-\frac{1}{x+3}=\frac{375}{376}\)
=>\(\frac{1}{x+3}=1-\frac{375}{376}\)
=>\(\frac{1}{x+3}=\frac{1}{376}\)
=>x+3=376
=>x=376-3
=>x=373
Vậy x=373
NM
4
4 tháng 8 2016
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\)\(\frac{1}{132}\)= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)\(=1-\frac{1}{12}=\frac{11}{12}\)
TK
1
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2021.2014}\)
\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2021}-\dfrac{1}{2014}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\left(1-\dfrac{1}{2014}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\dfrac{2013}{2014}=\dfrac{671}{2014}\)
\(B=\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+...+\dfrac{1}{2021\cdot2024}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2021\cdot2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{2024}\right)\\ =\dfrac{1}{3}\cdot\dfrac{2023}{2024}\\ =\dfrac{2023}{6072}\)