Tính: \(A=\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{19.20}}{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}\)
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Những câu hỏi liên quan
13 tháng 1 2018
Ta có: 1/1.2 = 1- 1/2
1/3.4 = 1/3 - 1/4
...............
1/19.20 = 1/19 - 1/20
Cộng vế với vế ta đc:
A = 1- 1/20 = 19/20
LA
13 tháng 1 2018
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{18}+\frac{1}{20}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{20}\right)\)
\(A=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)
\(\frac{A}{B}=1\)
Q
6 tháng 12 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(A=1-\frac{1}{20}\)
\(A=\frac{19}{20}\)
BT
6 tháng 12 2017
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}\)
\(=\frac{19}{20}\)
BD
1
29 tháng 1 2015
Ta có:A: 1/1.2 +1/2.3 +1/3.4+...+1/18.19+1/19.20
=> A= 1-1/2 +1/2-1/3+1/3-1/4+...+1/18-1/19+1/19-1/20
=>A= 1-1/20=19/20
HN
17 tháng 3 2018
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)
\(\Rightarrow x=50\)
Vậy x = 50
EG
17 tháng 3 2018
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)
\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)
28 tháng 6 2022
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
Tính tử số:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{20}-\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)
\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(\Rightarrow A=\frac{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}{\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}}=1\)