B=1.4+4.7+......+100+103
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DP
1
23 tháng 9 2023
`#3107`
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}\)
\(=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
`c)`
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
VN
Võ Ngọc Phương
CTVHS
VIP
23 tháng 9 2023
b) Sửa đề:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
VN
Võ Ngọc Phương
CTVHS
VIP
21 tháng 8 2023
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)
\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)
\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)
\(\dfrac{1}{x+3}=\dfrac{1}{103}\)
\(\Rightarrow x+3=103\)
\(x=103-3\)
\(x=100\)
Vậy x = 100
2
2
DN
24 tháng 8 2015
\(\frac{x}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{100.103}=\frac{102}{103}\)
\(\Leftrightarrow\frac{x-1}{1.4}+\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{100.103}\right)=\frac{102}{103}\)
\(\Leftrightarrow\frac{3\left(x-1\right)}{1.4}+\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{306}{103}\)
\(\Leftrightarrow\frac{3\left(x-1\right)}{1.4}+\frac{102}{103}=\frac{306}{103}\)
\(\Leftrightarrow\frac{3}{4}\left(x-1\right)=\frac{204}{103}\)
\(\Leftrightarrow x-1=\frac{272}{103}\)
\(\Leftrightarrow x=\frac{375}{103}\)
24 tháng 8 2015
=> 3x/4+3/4.7+3/7.10+...+3/100.103=306/103(nhân cả 2 vế của đt lên 2)
=>3x/4+(1/4-1/7)+(1/7-1/10)+...+(1/100-1/103)=306/103
=>3x/4+1/4-1/103+=306/103
=>3x/4+99/412=306/103
=>3x/4=306/103-99/412=1125/412
=>x=1125/412:3/4
=>x=1125/309
( nếu thấy đúng thì tick cho mk nha
12 tháng 5 2017
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...................+\dfrac{2}{97.100}\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{2}\left(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+..................+\dfrac{2}{97.100}\right)\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{97.100}\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+..............+\dfrac{1}{97}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{3}{2}A=1-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{3}{2}A=\dfrac{99}{100}\)
\(\Rightarrow A=\dfrac{99}{100}:\dfrac{3}{2}\)
\(\Rightarrow A=\dfrac{33}{50}\)
\(B=1.4+4.7+....+100.103\)
\(\frac{3}{B}=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{100.103}\)
\(\frac{3}{B}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{100}-\frac{1}{103}\)
\(\frac{3}{B}=\frac{1}{1}-\frac{1}{103}\)
\(\frac{3}{B}=\frac{102}{103}\)
\(B=\frac{3.103}{102}=\frac{103}{34}\)
thực ra dấu chấm là dấu nhân