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Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{18.19}+\frac{2}{19.20}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)\)
\(A=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)
\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\cdot\left(1-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{5}{6}\)
Hok tốt
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\)
=>\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=> 1-\(\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}=\frac{6}{6}+\frac{-1}{6}=\frac{5}{6}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{5}-\dfrac{1}{6}\)
\(A=1-\dfrac{1}{6}=\dfrac{5}{6}\)
\(\dfrac{1}{2}B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(\dfrac{1}{2}B=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(B=\dfrac{19}{20}\cdot2=\dfrac{19}{10}\)
Biết chu vi hình tròn là 14,444 đem.Tính diện tích hình tròn đó?
\(a.\) \(\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{1}-\dfrac{1}{6}\)
\(=\dfrac{6}{6}-\dfrac{1}{6}=\dfrac{5}{6}\)
\(b.\) \(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{18\cdot19}+\dfrac{2}{19\cdot20}\)
\(=\dfrac{2}{1}-\dfrac{2}{2}+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{18}-\dfrac{2}{19}+\dfrac{2}{19}-\dfrac{2}{20}\)
\(=\dfrac{2}{1}-\dfrac{2}{20}\)
\(=\dfrac{40}{20}-\dfrac{2}{20}\)
\(=\dfrac{38}{20}=\dfrac{19}{10}\)
a, Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
Hay \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(A=1-\dfrac{1}{6}\)
\(A=\dfrac{6}{6}-\dfrac{1}{6}\)
\(A=\dfrac{5}{6}\)
Vậy A = 5/6.
b, Đặt \(B=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{18.19}+\dfrac{2}{19.20}\)
\(B=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(B=2.\left(1-\dfrac{1}{20}\right)\)
\(B=2.\left(\dfrac{20}{20}-\dfrac{1}{20}\right)\)
\(B=2.\dfrac{19}{20}\)
\(B=\dfrac{19}{10}\)
Vậy B = 19/10.