Cho A = \(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
Hãy so sánh A với \(-\frac{1}{2}\)
K
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7 tháng 12 2016
A>1/2
Xin lỗi mình đang bận để lúc khác mình sẽ giải chi tiết
7 tháng 9 2018
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
3 tháng 5 2017
Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\left(-\frac{1.3}{2.2}\right).\left(-\frac{2.4}{3.3}\right)...\left(-\frac{99.101}{100.100}\right)\)
\(=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
Vậy \(A< -\frac{1}{2}\)
29 tháng 5 2017
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)
\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)
\(=\frac{101}{2\cdot100}\)
\(=\frac{101}{200}>\frac{1}{2}\)
26 tháng 4 2019
\(A=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(A=-\left(\frac{1.3}{2.2}\right).\left(\frac{2.4}{3.3}\right)....\left(\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2....99}{2.3...100}\right).\left(\frac{3.4....101}{2.3....100}\right)\)
\(A=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)
\(A=\frac{-101}{200}>\frac{-1}{2}\)
KN
13 tháng 5 2019
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{4}{2^2}\right)\left(\frac{1}{3^2}-\frac{9}{3^2}\right)...\left(\frac{1}{100^2}-\frac{10000}{100^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(\Rightarrow A=-\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\)
\(\Rightarrow A=-\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(\Rightarrow A=-\frac{\left(1.2...99\right)\left(3.4...101\right)}{\left(2.3...100\right)\left(2.3...100\right)}\)
\(\Rightarrow A=-\frac{101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
Vậy \(A< \frac{-1}{2}\)
\(A=\frac{-1.3}{2^2}.\frac{-2.4}{3^2}...\frac{-99.101}{100^2}\)
\(=-\left(\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\right)\)
\(=-\left(\frac{1}{100}.\frac{101}{2}\right)\)
\(=-\frac{101}{200}< \frac{-100}{200}=\frac{-1}{2}\)