K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
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
CTVHS
VIP
28 tháng 8 2023
\(\dfrac{3}{1.4}x+\dfrac{3}{4.7}x+\dfrac{3}{7.10}x+...+\dfrac{3}{31.34}x=33\)
\(x.3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{31.34}\right)=33\)
\(x.3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)=33\)
\(x.\left(1-\dfrac{1}{34}\right)=33\)
\(x.\dfrac{33}{34}=33\)
\(x=33:\dfrac{33}{34}=33.\dfrac{34}{33}\)
\(x=34\)
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
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
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
MIK CHẮC CHẮN 100% LÀ ĐÚNG, DẠNG TOÁN NÀY MIK LM NHIỀU R
HOK TỐT
27 tháng 4 2019
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{672}{2017}\)
\(\Rightarrow\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{672}{2017}\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{672}{2017}\cdot3=\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+3}=\frac{2017}{2017}-\frac{2016}{2017}\Rightarrow\frac{1}{x+3}=\frac{1}{2017}\)
\(\Rightarrow x+3=2017\Rightarrow x=2017-3\Rightarrow x=2014\)
C
2 tháng 3 2023
`3/1.4+3/4.7+3/7.10+...+3/94.97`
`=1/1-1/4+1/4-1/7+1/7-1/10+...+1/94-1/97`
`=1-1/97`
`=96/97`
2 tháng 3 2023
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)
O
25 tháng 3 2022
\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)
\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)
\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{16}\right)\)
\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)
\(=3\cdot\dfrac{15}{16}\)
\(=\dfrac{45}{16}\)
NV
25 tháng 4 2015
dễ mà
Gọi tổng đó là S. Theo đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}=\frac{42}{43}\)
= 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.