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DD
14 tháng 2 2022

\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{1830}+\frac{1}{1891}\)

\(=\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{3660}+\frac{2}{3782}\)

\(=\frac{2}{3\times4}+\frac{2}{4\times5}+\frac{2}{5\times6}+...+\frac{2}{60\times61}+\frac{2}{61\times62}\)

\(=\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+...+\frac{2}{60}-\frac{2}{61}+\frac{2}{61}-\frac{2}{62}\)

\(=\frac{2}{3}-\frac{2}{62}=\frac{59}{93}\)

29 tháng 1 2023

=2x(1/12+1/20+1/30+...+1/3660+1/3782)

=2x(1/(3x4)+1/(4x5)+1/(5x6)+...+1/(60x61)+1/(61x62)

=2x(1/3-1/4+1/4-1/5+1/5-1/6+...+1/60-1/61+1/61-1/62)

=2x(1/3-1/62)=2x59/186=59/93

15 tháng 6 2023

3 x 15 + 21 x 15 + 85 x 5

= 45 + 315 + 425

= 785

15 - 30 + 40 

= 25

21 + 19 - 50 + 10

= 0

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=-\dfrac{1}{20}+2\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{12}\times\dfrac{3}{12}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=-\dfrac{9}{20}\)

15 tháng 6 2023

\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)

\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)

\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)

\(\dfrac{1}{5}-\dfrac{1}{4}+2\)

\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)

\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)

\(=\dfrac{44}{20}-\dfrac{5}{20}\)

\(=\dfrac{39}{20}\)

\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{12}\times\dfrac{1}{4}\)

\(=\dfrac{5}{48}\)

\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)

\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)

\(=\dfrac{6}{20}-\dfrac{15}{20}\)

\(=-\dfrac{9}{20}\)

 

18 tháng 4 2019

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(\frac{A}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)

\(\frac{A}{2}=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{12}\)

\(\Rightarrow A=\frac{2}{4}-\frac{2}{12}=\frac{16}{48}\)

\(B=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)

\(\frac{B}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)

\(\frac{B}{2}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow B=\frac{2}{3}-\frac{2}{11}=\frac{16}{33}\)

Mà \(\frac{16}{48}< \frac{16}{33}\Rightarrow A< B\)

Vậy : A < B

24 tháng 9 2018

16/21 nha

27 tháng 8 2023

\(C\text{=}\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}\)

\(C\text{=}\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{10}+\dfrac{1}{15}\right)+\left(\dfrac{1}{21}+\dfrac{1}{28}\right)\)

\(C\text{=}\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}\)

\(C\text{=}\dfrac{3}{4}\)

23 tháng 5 2023

S1/2=1/2+1/6+1/12+...+1/600

S1/2=1/1*2+1/2*3+....+1/24*25

S1/2=1-1/25

S1/2=24/25

S=48/25

19 tháng 5 2023

A =          1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +  \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\)

A = 2\(\times\) ( \(\dfrac{1}{2}\)  +  \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) +  \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\)\(\dfrac{1}{72}\))

A =2\(\times\)\(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\))

A = 2 \(\times\) ( \(\dfrac{1}{1}\)\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))

 A = 2\(\times\)( 1 - \(\dfrac{1}{9}\))

A = 2 \(\times\) \(\dfrac{8}{9}\)

A = \(\dfrac{16}{9}\)

AH
Akai Haruma
Giáo viên
9 tháng 11 2021

Lời giải:

$\frac{A}{2}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$
$=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}+\frac{9-8}{8\times 9}+\frac{10-9}{9\times 10}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}$

$=1-\frac{1}{9}=\frac{8}{9}$

$\Rightarrow A=2\times \frac{8}{9}=\frac{16}{9}$

23 tháng 5 2021

 B =1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28 

 B = 1 - 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 - 1/28 

 B = 1 - ( 1/3 + 1/3 - 1/6 + 1/6 - 1/10 + 1/10 - 1/15 + 1/15 - 1/21 + 1/21 ) - 1/28

  B = 1 - 1/28

  B  = 27/28

~ Hok T ~

8 tháng 8 2018

nhớ giải chi tiết

9 tháng 8 2018

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{105}+\frac{1}{210}\)

 =>   \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{210}+\frac{1}{240}\)

                 \(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{14.15}+\frac{1}{15.16}\)

                 \(=\frac{1}{2}-\frac{1}{3}+\frac{!}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

               \(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)

=>   \(A=\frac{7}{8}\)