So Sánh A = 124 . [1/(1.1985) + 1/(2.1986) +....+ 1/(16.2000)
Và B = 1/(1.17) + 1/(2.18) + 1/(3.19) +.....+ 1/(1984.2000)
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YN
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`
CC
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: \(B=\frac{1}{16}\cdot\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+\frac{1}{3}-\frac{1}{19}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}-\frac{1}{17}-\frac{1}{18}-\frac{1}{19}-...-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2000}\right)\right]\)