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VN
Võ Ngọc Phương
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
13 tháng 5 2015
9A = 1.4.[7+2] + 4.7. [10-1] + 7.10.[13-4] +...+ 91.94. [97-88]
= 1.4.7 + 1.2.4 + 4.7.10 - 1.4.7 + 7.10.13 - 4.7.10+...+ 91.94.97 - 88.91.94
= 1.2.4 + 91.94.97 = 8 +829738 = 829746 => A = 829746 : 9 = 92194
đúng cái nhé
CD
2
DL
29 tháng 6 2017
Đặt : \(A=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{27\cdot30}\)
\(A=\frac{1}{3}\left(\frac{5}{1}-\frac{5}{4}+\frac{5}{4}-\frac{5}{7}+...+\frac{5}{27}-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\left(5-\frac{5}{30}\right)\)
\(A=\frac{1}{3}\cdot\frac{29}{6}\)
\(A=\frac{29}{18}\)
DP
29 tháng 6 2017
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+....+\frac{5}{27.30}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{30-27}{27.30}\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}\cdot\frac{29}{30}=\frac{29}{18}\)
8 tháng 6 2018
f,F=3. (1/2 .3 + 1/3.4 +...+ 1/99.100)
= 3. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
= 3. (1/2 - 1/100)
= 3. 49/100
= 147/100
g, G = 5/3. (3/1.4 + 3/4.7 +...+ 3/61.64)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 +...+ 1/61 - 164
= 5/3 . (1-1/64)
= 5/3 . 63/64
= 105/64
MX
8 tháng 6 2018
f, \(F=\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(\Leftrightarrow F=3\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{49}{100}\right)=\frac{147}{100}\)
g, \(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Leftrightarrow G=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{61.64}\right)\)
\(\Leftrightarrow G=5.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}\left(1-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}.\frac{63}{64}=\frac{105}{64}\)
8 tháng 6 2018
\(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Rightarrow G=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{61.64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{64}\right)=\frac{5}{3}.\frac{63}{64}\)
\(\Rightarrow G=\frac{5.63}{3.64}=\frac{5.21.3}{3.64}=\frac{5.21}{64}=\frac{105}{64}\)
29 tháng 6 2017
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
DL
1 tháng 7 2017
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
BD
3
19 tháng 3 2018
Ta có :
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
Vậy \(A=\frac{33}{50}\)
Chúc bạn học tốt ~
T
19 tháng 3 2018
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
PQ
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 8 2023
Để olm.vn giúp em nhá
C = \(\dfrac{1}{1.4}\) + \(\dfrac{1}{4.7}\) + \(\dfrac{1}{7.11}\)+...+ \(\dfrac{1}{994.997}\) + \(\dfrac{1}{997.1000}\)
C = \(\dfrac{1}{3}\).( \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\)+...+ \(\dfrac{3}{994.997}\)+ \(\dfrac{3}{997.1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+...+ \(\dfrac{1}{994}\)- \(\dfrac{1}{997}\)+ \(\dfrac{1}{997}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{1000}\))
C = \(\dfrac{1}{3}\). \(\dfrac{999}{1000}\)
C = \(\dfrac{333}{1000}\)
NT
2
8 tháng 7 2015
\(E=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{97\cdot100}\)
\(=\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right):3\)
\(=\left(1-\frac{1}{100}\right):3=\frac{33}{100}\)
\(F=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{74\cdot101}\)
\(=\left(3-\frac{3}{5}+\frac{3}{5}-\frac{3}{9}+...+\frac{3}{74}-\frac{3}{101}\right):4\)
\(=\left(3-\frac{3}{101}\right):4=\frac{75}{101}\)
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/31.34
= 1/3 . ( 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/31.34 )
= 1/3 . ( 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/31 - 1/34 )
= 1/3 . ( 1 - 1/34 )
= 1/3 . 33/34
= 11/34