\(\frac{1}{10}\) +\(\frac{1}{40}+...">
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Ta có: \(A=\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}\)

\(=\dfrac{1}{2\text{x}5}+\dfrac{1}{5\text{x}8}+...+\dfrac{1}{17\text{x}20}\)

\(=\dfrac{1}{3}\text{x}\left(\dfrac{3}{2\text{x}5}+\dfrac{3}{5\text{x}8}+...+\dfrac{3}{17\text{x}20}\right)\)

\(=\dfrac{1}{3}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{3}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\text{x}\dfrac{9}{20}=\dfrac{3}{20}\)

20 tháng 3

Gợi ý nhé 10 = 2.5, 40 = 5.8, 88 = 8 . 11 , 154 = 11 . 14 , 238 = 14 . 17 , 340 = 17 . 20

2 tháng 9 2015

Ta có:\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{340}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)

=        \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{17}-\frac{1}{20}\right)=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

2 tháng 4 2017

\(=\frac{3}{20}=0,15\)

11 tháng 6 2018

Ta có : 

\(N=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(N=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3N=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(3N=\frac{1}{2}-\frac{1}{20}\)

\(3N=\frac{9}{20}\)

\(N=\frac{9}{20}:3\)

\(N=\frac{3}{20}\)

Vậy \(N=\frac{3}{20}\)

Chúc bạn học tốt ~ 

11 tháng 6 2018

\(N=\frac{1}{10}+\frac{1}{40}+...+\frac{1}{238}+\frac{1}{340}\)

\(N=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{14.17}+\frac{1}{17.20}\)

\(N=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(N=\frac{1}{2}-\frac{1}{20}\)

\(N=\frac{10}{20}-\frac{1}{20}\)

\(N=\frac{9}{20}\)

17 tháng 6 2017

\(\Rightarrow\left[\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\right]\cdot\frac{2x}{10}\)

\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)

\(\Rightarrow\left[\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]\right]\cdot20=\frac{2x}{10}\)

\(\Rightarrow\left[\frac{1}{3}\cdot\frac{9}{20}\right]\cdot20=\frac{2x}{10}\)

\(\Rightarrow\frac{3}{20}\cdot20=\frac{2x}{10}\)

\(\Rightarrow3\cdot20=\frac{2x}{10}\Leftrightarrow60=\frac{2x}{10}\)

=> 2x = 60*10

=> 2x = 600

=> x = 300

17 tháng 6 2017

\(\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\right).20=\frac{2x}{10}\)

\(\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\right).20=\frac{2x}{10}\)

\(\left[3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)

\(\left[3.\left(\frac{1}{2}-\frac{1}{20}\right)\right].20=\frac{2x}{10}\)

\(\left(3.\frac{9}{20}\right).20=\frac{2x}{10}\)

\(\frac{27}{20}.20=2x\div10\)

\(27=2x\div10\)

\(x=27\times10\div2\)

\(\Rightarrow x=135\)

29 tháng 6 2017

Toán quá dễ. Tự túc là hạnh phúc mọi nhà bn nhé !

29 tháng 6 2017

\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)

Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\)\(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có : 

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)

\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)

\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)

\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)

2 tháng 5 2017

A=1/3x(1/2x5+1/5x8+......+1/20x23)

A=1/3x(1/2-1/5+1/5-1/8+......+1/20-1/23)

A=1/3x(1/2-1/23)

A=1/3x21/46

A=7/46

2 tháng 1 2018

\(1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times1\frac{1}{6}\times1\frac{1}{7}\times1\frac{1}{8}\)

\(=\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times\frac{9}{8}\)

\(=\frac{9}{3}\)

\(=3\)

2 tháng 1 2018

 = 4/3 x 5/4 x 6/5 x 7/6 x 8/7 x 9/8

 =  4x5x6x7x8x9/3x4x5x6x7x8 = 9/3 = 3

k mk nha

7 tháng 8 2017

\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)

\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)

\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)

\(=1+\frac{28}{5}\)

\(=\frac{33}{5}\)

7 tháng 8 2017

Ta có:

a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)

b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)

\(=1+\frac{31}{128}=1\frac{31}{128}\)

11 tháng 3 2017

=3/2*4/3*5/4...*11/10

=5,5

14 tháng 5 2017

Gọi tổng trên là A

Ta có : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{2560}\)

\(2A=2\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(2A=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)

\(2A-A=\left(\frac{2}{5}+\frac{1}{5}+...+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(A\left(2-1\right)=\frac{2}{5}-\frac{1}{2560}\)

\(A.1=\frac{1024}{2560}-\frac{1}{2560}\) 

\(A=\frac{1023}{2560}\)

14 tháng 5 2017

Ta có : A = 1/5 + 1/10 + 1/20 + ... + 1/2560

2A = 2 ( 1/5 + 1/10 + ... + 1/2560 )

2A = 2/5 + 1/5 + 1/10 + .. + 1/2560

2A - A = ( 2/5 + 1/5 + ... + 1/1280 ) - ( 1/5 + 1/10 + ... + 1/2560 )

A =  2 - 1 = 2/5 - 1/2560

A.1 = 1024/2560 - 1/2560

A = 1023 = 2560