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3 tháng 2 2017
ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)
3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)
3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)
3A=\(\frac{1}{5}-\frac{1}{608}\)
3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)
3 tháng 2 2017
Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)
3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)
3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)
3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)
A=\(\frac{201}{3040}\)
6 tháng 8 2020
(x+1/4-1/3).(13/6-1/4)=7/46
(x+1/4-1/3).23/12=7/46
(x+1/4-1/3)=7/46:23/12
(x+1/4-1/3)=7/46.12/23
(x+1/4-1/3)=42/529
x+1/4=42/529+1/3
x+1/4=655/1587
x=655/1587-1/4
x=1033=/6348
vậy x=1033/6348
KL
8 tháng 4 2016
a.2/3 + 1/3 - x= 3/5
=> 1-x = 3/5
=> x = 1-3/5= 2/5
b. 1/8/15 - 2/3x = 0,2
=> 2/3x= 23/15 - 1/5= 4/3
=> x= 4/3 : 2/3=2
c. 2/3x -3/2x = 5/12
=> x( 2/3 - 3/2) = 5/12
=> x. -5/6 = 5/12
=> x= 5/12 : -5/6
=> x= -1/2
8 tháng 4 2016
a) 2/3 + 1/3 - x = 3/5
=> 1 - x = 3/5
=> x = 1 - 3/5
x = 2/5
b) \(1\frac{8}{15}-\frac{2}{3}x=0,2\)
=> 23/15 - 2/3x = 0,2
=> 2/3x = 23/15 - 0,2
2/3x = 1,333333333
=>x = 1,333333333 : 2/3
x = 2
c) ???
22 tháng 6 2020
a, \(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}\)
\(=\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{7}{21}\)
\(=0+\frac{7}{21}\)
\(=\frac{7}{21}\)
\(=\frac{1}{3}\)
b, \(\frac{8}{9}+\frac{1}{9}.\frac{7}{9}+\frac{1}{9}.\frac{2}{9}\)
\(=\frac{8}{9}+\frac{1}{9}.\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{8}{9}+\frac{1}{9}.1\)
\(=\frac{8}{9}+\frac{1}{9}\)
\(=1\)
22 tháng 6 2020
a) \(\frac{-3}{5}\)+\(\frac{7}{21}\)+\(\frac{-4}{5}\)+\(\frac{7}{5}\)
=(\(\frac{-3}{5}\)+\(\frac{-4}{5}\)+\(\frac{7}{5}\)) +\(\frac{7}{21}\)
= 0+
3 tháng 2 2019
\(G=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..............+\frac{1}{3^{100}}\)
\(3G=1+\frac{1}{3}+\frac{1}{3^2}+...............+\frac{1}{3^{99}}\)
\(3G-G=\left(1+\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...............+\frac{1}{3^{100}}\right)\)
\(2G=1-\frac{1}{3^{100}}\)
\(\Rightarrow G=\left(1-\frac{1}{3^{100}}\right):2\)
S
18 tháng 3 2017
a, \(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 3
x = 305
b, \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{1}{2}.\frac{3984}{1993}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{1992}{1993}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)
=> x + 1 = 1993
x = 1993 - 1
x = 1992
N
18 tháng 3 2017
a ,\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
KN
15 tháng 4 2019
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
KN
15 tháng 4 2019
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
29 tháng 6 2022
a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
Bằng \(\frac{1708}{625}\)
= 1708/625
k mình nha