Tính: A=\((1-\frac{1}{2^2})\cdot(1-\frac{1}{3^2})\ast\ast\ast(1-\frac{1}{50^2})\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TV
17 tháng 4 2019
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
DT
6 tháng 4 2019
đề thiếu bn ơi
phải là \(\frac{1}{99^2}-1\)
A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{98^2}-1\right)\).\(\left(\frac{1}{99^2}-1\right)\)
do tích A có: (99-2)+1=98 thừa số nguyên âm nên tích A dương
A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)=\(\frac{1.3.2.4.3.5...97.99.98.100}{2^2.3^2.4^2...98^2.99^2}\)
=\(\frac{1.2.3.4...98}{2.3.4...98.99}.\frac{3.4.5...99.100}{2.3.4...98.99}\)=\(\frac{1}{99}.\frac{100}{2}\)=\(\frac{50}{99}\)
vậy A=\(\frac{50}{99}\)
#HỌC TỐT#
17 tháng 10 2018
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
10 tháng 3 2017
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{32}\left(1+2+3+...+32\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{32}.\frac{32.\left(32+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{32+1}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{33}{2}\)
\(\frac{2+3+4+....+33}{2}\)
\(=\frac{\frac{33\left(33+1\right)}{2}-1}{2}=280\)
23 tháng 4 2016
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
23 tháng 4 2016
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
A=\(\frac{1.3.2.4.\ldots49.51}{2.2.3.3\ldots50.50}=\frac{51}{50}.\)
Ta có: \(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{50^2}\right)\)
\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{50}\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{50}\right)\)
\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{49}{50}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{51}{50}=\frac{1}{50}\cdot\frac{51}{2}=\frac{51}{100}\)