Tính giá trị các biểu thức sau
A=\(\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2015.2017}\right)\)
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Những câu hỏi liên quan
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
HY
3 tháng 4 2017
Ta có:
\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2016^2}{2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{2016^2}{2015.2017}\)
\(A=\left(\dfrac{2.3.4...2016}{2.3.4.5...2015}\right).\left(\dfrac{2.3.4...2016}{2.3.4.5...2017}\right)\)
\(A=2016.\dfrac{1}{2017}=\dfrac{2016}{2017}\)
AH
Akai Haruma
Giáo viên
2 tháng 1 2019
Lời giải:
Xét tổng quát:
\(1+\frac{1}{k(k+2)}=\frac{k(k+2)+1}{k(k+2)}=\frac{(k+1)^2}{k(k+2)}\)
Thay $k=1,2,....,2015$ ta có:
\(1+\frac{1}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{4^2}{3.5}\)
\(1+\frac{1}{4.6}=\frac{5^2}{4.6}\)
.............
\(1+\frac{1}{2015.2017}=\frac{2016^2}{2015.2017}\)
Nhân theo vế:
\(\Rightarrow A=\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}....\frac{2016^2}{2015.2017}\)
\(=\frac{(1.2.3...2016)^2}{(1.2.3...2015)(2.3.4...2017)}=\frac{(1.2.3...2016)(2.3....2016)}{(1.2.3...2015)(2.3.4...2017)}=2016.\frac{1}{2017}=\frac{2016}{2017}\)
11 tháng 8 2017
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2015.2017}\right)\)\(A=\dfrac{1}{2}.\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{4064256}{4064255}\)
\(A=\dfrac{1}{2}.\dfrac{2.2.3.3.4.4....2016.2016}{3.8.15....4064255}\)
\(A=\dfrac{1}{2}.\dfrac{2.2.3.3.4.4....2016.2016}{1.3.2.4.3.5.....2015.2017}\)
\(A=\dfrac{1}{2}.\dfrac{2.3.4....2016}{1.2.3.....2015}.\dfrac{2.3.4.....2016}{3.4.5.....2017}\)
\(A=\dfrac{1}{2}.2016.\dfrac{2}{2017}\)
\(A=1008.\dfrac{2}{2017}\)
\(A=\dfrac{2016}{2017}\)
9 tháng 3 2021
\(A=\dfrac{1}{2}\left(2.\dfrac{2}{3}\right)\left(\dfrac{3}{2}.\dfrac{3}{4}\right)\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
