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8 tháng 11 2017

Mk chịu thui =)) Sorry ^o^

6 tháng 4 2022

`Answer:`

\(A=124.\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)

\(=\frac{124}{1984}.\left(\frac{1984}{1.1985}+\frac{1984}{2.1986}+\frac{1984}{3.1987}+...+\frac{1984}{16.2000}\right)\)

\(=\frac{1}{16}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

\(=\frac{1}{16}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\)

\(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)

\(=\frac{1}{16}.\left(\frac{16}{1.17}+\frac{16}{2.18}+...+\frac{16}{1984.2000}\right)\)

\(=\frac{1}{16}.\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)

\(=\frac{1}{16}.\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)

\(=\frac{1}{16}.\left(1+\frac{1}{2}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{1984}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\)

\(=\frac{1}{16}.[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)]\)

`=>A=B`

21 tháng 7 2019

\(A=124\left(\frac{1}{1.1985}+\frac{1}{2.1986}+\frac{1}{3.1987}+...+\frac{1}{16.2000}\right)\)

      \(=\frac{124}{1984}.\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

       \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

  Và \(B=\frac{1}{1.17}+\frac{1}{2.18}+...+\frac{1}{1984.2000}\)

             \(=\frac{1}{16}\left[\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+...+\frac{1}{1984}-\frac{1}{2000}\right)\right]\)

              \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{2000}\right)\right]\)

=      \(\frac{1}{16}\)  .    \(\left[\left(1+...+\frac{1}{16}\right)+\left(\frac{1}{17}+...+\frac{1}{1984}-\frac{1}{17}-...-\frac{1}{1984}\right)-\left(\frac{1}{1985}+...+\frac{1}{2000}\right)\right]\)

 = \(=\frac{1}{16}\left[\left(1+\frac{1}{2}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+...+\frac{1}{2000}\right)\right]\)

Vậy A = B

3 tháng 11 2017

Ta có: \(A=124\cdot\frac{1}{1984}\cdot\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)

\(\Rightarrow A=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\right]\)

Laji cos: \(B=\frac{1}{16}\cdot\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+\frac{1}{3}-\frac{1}{19}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)

\(\Rightarrow B=\frac{1}{16}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}-\frac{1}{17}-\frac{1}{18}-\frac{1}{19}-...-\frac{1}{2000}\right)\)

\(\Rightarrow B=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2000}\right)\right]\)