NN
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
14 tháng 1 2024
\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\cdot...\left(1+\dfrac{1}{2499}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot...\cdot\dfrac{2500}{2499}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{50\cdot50}{49\cdot51}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot50}{1\cdot2\cdot3\cdot...\cdot49}\cdot\dfrac{2\cdot3\cdot...\cdot50}{3\cdot4\cdot...\cdot51}\)
\(=\dfrac{50}{1}\cdot\dfrac{2}{51}=\dfrac{100}{51}\)
NT
0
D
22 tháng 4 2015
a)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\) \(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{49.51}{50^2}\)\(=\frac{\left(2.3.4.....49\right)\left(4.5.6....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}=\frac{2.51}{50.3}=\frac{1.17}{25}=\frac{17}{25}\)
D
22 tháng 4 2015
b)\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
HN
0
24 tháng 2 2020
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{16}{15}\right)...\left(1+\frac{2500}{2499}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{2500}{2499}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{50.50}{49.51}\)
\(=2.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}...\frac{50}{49}.\frac{50}{51}\)
\(=\left(2.\frac{3}{2}.\frac{4}{3}...\frac{50}{49}\right)\left(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{50}{51}\right)\)
\(=\frac{2.3.4...50}{2.3.4...49}.\frac{2.3.4...50}{3.4.5...51}\)
\(=50.\frac{2}{51}=\frac{100}{51}\)
6 tháng 5 2017
\(A=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.....1\frac{1}{360}\)
\(A=1+\left(\frac{1}{3}.\frac{1}{8}.\frac{1}{15}.\frac{1}{24}.....\frac{1}{360}\right)\)
Nếu đúng thì tk nha
S
30 tháng 7 2016
Ta có: 1/3 ; 1/15 ; 1/35;...
<=> 1/1.3 ; 1/3.5 ; 1/5.7
=> chữ số thứ 100 là: 1/199.201
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
A = \(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)\)\(...\left(1+\frac{1}{2499}\right)\)
A = \(\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{8}{8}+\frac{1}{8}\right)\left(\frac{15}{15}+\frac{1}{15}\right)\)\(...\left(\frac{2499}{2499}+\frac{1}{2499}\right)\)
A = \(\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{2500}{2499}\)
A = \(\frac{4.9.16.....2500}{3.8.15.....2499}\)
A = \(\frac{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(50.50\right)}{3.8.15.24.....2499}\)
A = \(\frac{2.3.4.....50}{3.4.5.6.....51}\)
A = \(\frac{2}{51}\)
Vậy A = \(\frac{2}{51}\)
( Nếu sai mong bạn thông cảm ạ ! )
_HT_
Answer:
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{2499}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{2500}{2499}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}\)
\(=\frac{2^2.3^2.4^2...50^2}{1.3.2.4.3.5...49.51}\)
\(=\frac{2.50}{51}\)
\(=\frac{100}{51}\)