LP
1
K
Khách
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NV
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
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
AN
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=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)
7 tháng 2 2023
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
Q
30 tháng 3 2017
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{55}+\dfrac{1}{66}\)
\(A=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\right)\)
\(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\right)\)
\(A=2\left(\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{12}\right)\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\Rightarrow A=\dfrac{1}{3}\)
LH
1
NP
6 tháng 5 2015
a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
Đặt \(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)
\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)
\(\Rightarrow C=\frac{1}{6}\cdot2\)
\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)
NG
1
NM
24 tháng 3 2019
a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{14}{30}=\frac{7}{15}\)
LN
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)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)
\(...\)
PT
1
27 tháng 4 2016
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{132}\)
\(A=\frac{1}{2}.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{12}\right)\)
\(A=\frac{1}{2}.\frac{1}{6}\)
\(A=\frac{1}{12}\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(\frac{A}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(\frac{A}{2}=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)
\(\frac{A}{2}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(\frac{A}{2}=\frac{1}{4}-\frac{1}{12}\)
\(\Rightarrow A=\frac{2}{4}-\frac{2}{12}=\frac{16}{48}\)
\(B=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)
\(\frac{B}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)
\(\frac{B}{2}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(\frac{B}{2}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{B}{2}=\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow B=\frac{2}{3}-\frac{2}{11}=\frac{16}{33}\)
Mà \(\frac{16}{48}< \frac{16}{33}\Rightarrow A< B\)
Vậy : A < B