3/1.4 + 3/4.7 + 3/7.10 + ... + 3/ x. (x + 3)
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28 tháng 8 2023
\(\dfrac{3}{1\times4}x+\dfrac{3}{4\times7}x+\dfrac{3}{7\times10}x+...+\dfrac{3}{31\times34}x=33\)
\(x\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{31\times34}\right)=33\)
\(x\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x\left(1-\dfrac{1}{34}\right)=33\)
\(\dfrac{33}{34}x=33\)
\(x=34\)
VN
Võ Ngọc Phương
VIP
28 tháng 8 2023
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
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`
8 tháng 8 2018
Làm từng phần nha bạn
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)
Đặt \(A+x=\frac{299}{301}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)
\(A=1-\frac{1}{301}\)
\(A=\frac{300}{301}\)
=> \(\frac{300}{301}+x=\frac{299}{301}\)
\(x=\frac{299-300}{301}\)
\(x=-\frac{1}{301}\)
8 tháng 8 2018
\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)
\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=\frac{303}{304}\)
\(A=\frac{505}{304}\)
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
NH
1
LH
16 tháng 7 2015
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{x+3}\right)=\frac{667}{2002}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{667}{2002}:\frac{1}{3}\)
\(\frac{1}{1}-\frac{1}{x+3}=\frac{2001}{2002}\)
\(\frac{1}{x+3}=1-\frac{2001}{2002}\)
\(\frac{1}{x+3}=\frac{1}{2002}\)
\(\frac{1}{x}=\frac{1}{2002-3}\)
\(\frac{1}{x}=\frac{1}{1999}\)
Vậy x = 1999
TN
14 tháng 5 2016
đặt VT là A ta có:
\(3A=3\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}\right)\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{6}{19}\)
\(3A=1-\frac{1}{x+3}\)
\(A=\left(1-\frac{1}{x+3}\right):3\)
thay A vào VT ta đc:\(\left(1-\frac{1}{x+3}\right):3=\frac{6}{19}\)
\(1-\frac{1}{x+3}=\frac{18}{19}\)
\(\frac{1}{x+3}=\frac{1}{19}\)
=>x+3=19
=>x=16
NH
0
DY
3
27 tháng 4 2019
1/ 1.4+ 1/ 4.7+ 1/ 7.10+....+1/ x.( x+ 3)= 672/ 2017
(3/1.4+3/4.7+3/7.10+...+ 3/x(x+3)).1/3=672/2017
(1/1-1/4+1/4-1/7+1/7-1/10+.....+(x+3)-x/x.(x+3)).1/3=672/2017
(1/1-1/(x+3)).1/3=672/2017
1/1-1/(x+3)= 672/2017:1/3
1/1-1/(x+3) = 2016/2017
1/(x+3)=1/1-2016/2017
1/(x+3)=1/2017
x+3=2017
x= 2017-3
x= 2014
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HOK TỐT
27 tháng 4 2019
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{672}{2017}\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}\cdot3=\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+3}=\frac{2017}{2017}-\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=\frac{1}{2017}\)
\(\Rightarrow x+3=2017\Rightarrow x=2017-3\Rightarrow x=2014\)
= 1-1/4+1/4-1/7+1/7-1/10 +...+ 1/x -1/ x+3
= 1 -1/x+3
= x+2 / x+3
Câu này dễ mà.