Tính:
(1-1/3).(1-1/6).(1-1/10)...(1-1/210)
K
Khách
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
16 tháng 4 2021
Lời giải:
$\frac{A}{2}=\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-...-\frac{1}{420}$
$\frac{A}{2}=\frac{1}{2}-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)$
Xét:
$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}$
$=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}$
$=\frac{1}{2}-\frac{1}{21}$
Do đó:
$\frac{A}{2}=\frac{1}{2}-(\frac{1}{2}-\frac{1}{21})=\frac{1}{21}$
$\Rightarrow A=\frac{2}{21}$
K
1
20 tháng 4 2016
Mình tính đc kết quả 22 nhưng hơi khó hiểu mong mọi người giải dùm
K
6
TA
22 tháng 4 2016
đặt A=(1-1/3)........
Ta có A=\(\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}\cdot...\cdot\frac{209}{210}=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}\cdot...\cdot\frac{418}{420}=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{19\cdot22}{20\cdot21}\)
=\(\frac{1\cdot4\cdot2\cdot5\cdot3\cdot6\cdot...\cdot19\cdot22}{2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot...\cdot20\cdot21}=\frac{\left(1\cdot2\cdot3\cdot...\cdot19\right)\cdot\left(4\cdot5\cdot6\cdot...\cdot22\right)}{\left(2\cdot3\cdot4\cdot...\cdot20\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot21\right)}\)
=\(\frac{1\cdot22}{20\cdot3}=\frac{11}{30}\)
LC
22 tháng 4 2016
Đặt \(A=\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)
=>\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)
=>\(A=\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}.\frac{14.2}{15.2}...\frac{209.2}{210.2}\)
=>\(A=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{418}{420}\)
=>\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{19.22}{20.21}\)
=>\(A=\frac{\left(1.4\right).\left(2.5\right).\left(3.6\right).\left(4.7\right)...\left(19.22\right)}{\left(2.3\right).\left(3.4\right).\left(4.5\right).\left(5.6\right)...\left(20.21\right)}\)
=>\(A=\frac{\left(1.2.3.4...19\right).\left(4.5.6.7...22\right)}{\left(2.3.4.5...20\right).\left(3.4.5.6...21\right)}\)
=>\(A=\frac{1.22}{20.3}\)
=>\(A=\frac{22}{60}=\frac{11}{30}\)
Vậy \(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)=\frac{11}{30}\)
LT
0
DQ
0
HT
7
DL
9 tháng 8 2017
Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{100}{101}\cdot\frac{1}{2}=\frac{50}{101}\)
TP
9 tháng 8 2017
Ta có:
a)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
b)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{210}\)
\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)
XO
3 tháng 3 2020
C = \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{210}\right)=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{209}{210}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}=\frac{2.2\left(2.3.4...19\right)\left(5.6...22\right)}{\left(2.3.4..20\right)\left(3.4.5..21\right)}=\frac{4.22}{19.3.4}=\frac{22}{57}\)
B
30 tháng 3 2019
\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)
\(B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{3}.\frac{1}{3}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{6}.\frac{9}{2}.\frac{14}{15}...\frac{209}{210}\)
\(B=\frac{1}{2}.\frac{1}{1}.\frac{7}{5}...\frac{209}{210}\)
\(B=\frac{7}{10}...\frac{209}{210}\)
\(B=\frac{62}{210}\)
DP
14 tháng 8 2017
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
DP
14 tháng 8 2017
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)