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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

8 tháng 5 2023

do những số đó bé hơn 1 nên cộng lại vẫn bé hơn 1

 

11 tháng 5 2023

  A =  \(\dfrac{1}{3}\) +    \(\dfrac{1}{6}\) +  \(\dfrac{1}{10}\)  + \(\dfrac{1}{15}\) + ..+ \(\dfrac{1}{55}\)\(\dfrac{1}{66}\)

A  = 2  \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\)  + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) +...+ \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\))

A  = 2 \(\times\) ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) +  \(\dfrac{1}{4.5}\)\(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{10.11}\)\(\dfrac{1}{11.12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)\(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))

A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))

A = 1 - \(\dfrac{1}{6}\) < 1

Vậy A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{55}\)\(\dfrac{1}{66}\) < 1 

  

10 tháng 4 2017

A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)

A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)

A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)

10 tháng 4 2017

tớ làm hơi gọn nên có gì kho hiểu thì nói tớ

a: -15/37>-25/37

b: -13/21=-26/42

-9/14=-27/42

mà -26>-42

nên -13/21>-9/14

c: -49/-63=7/9

56/80=7/10

=>-49/-63>56/80

d: 3/14=1-11/14

4/15=1-11/15

mà 11/14>11/15

nên 3/14<4/15

30 tháng 3 2017

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{55}+\dfrac{1}{66}\)

\(A=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\right)\)

\(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\right)\)

\(A=2\left(\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{12}\right)\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\Rightarrow A=\dfrac{1}{3}\)

6 tháng 5 2015

a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)

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

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

\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)

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

\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)

\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)

\(\Rightarrow C=\frac{1}{6}\cdot2\)

\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)

 

20 tháng 10 2017

BẠN NÀO BIẾT THÌ GIÚP MÌNH VỚI !!!!!!!

24 tháng 3 2019

a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

\(=\frac{14}{30}=\frac{7}{15}\)

24 tháng 3 2019

a)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=2\left(1-\frac{1}{15}\right)\)

\(=2.\frac{14}{15}\)

\(=\frac{28}{15}\)

b)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)

                                                                                         \(...\)

24 tháng 12 2020

a

nAK.DNX. 0pwi9dOjkciopjopoijasd

27 tháng 4 2016

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

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

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

\(A=\frac{1}{2}.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\right)\)

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

\(A=\frac{1}{2}.\frac{1}{6}\)

\(A=\frac{1}{12}\)