Cho A= (1/2^2-1) (1/3^2-1) (1/4^2-1) ... (1/2013^2-1) (1/2014^2-1) và B=-1/2. Hãy so sánh A và B.
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17 tháng 8 2021
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
1 tháng 7 2019
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)
\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)
\(A=\frac{-2015}{4028}\)
QH
Cho A = (1/2^2 - 1)(1/3^2 - 1) (1/4^2 - 1) ... (1/2013^2 -1)(1/2014^2 - 1) Và B = -1/2
So sánh A và B
0
20 tháng 8 2021
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
20 tháng 8 2021
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
TT
10 tháng 8 2020
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(\Leftrightarrow A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{4052169}-1\right)\left(\frac{1}{\text{}\text{}4056196}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-4056195}{\text{}4056196}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3}{2^2}.\frac{\left(-2\right)4}{3^3}.\frac{\left(-3\right)5}{4^2}.....\frac{\left(-2013\right)2015}{\text{}2014^2}\)
\(\Leftrightarrow A=\frac{\left(-1\right)\left(-2\right)....\left(-2013\right)}{2.3...1014}.\frac{3.4......2015}{2.3......2014}\)
\(\Leftrightarrow A=\frac{-1}{1014}.\frac{2015}{2}=\frac{-2015}{4028}\)
VÌ \(\frac{-2015}{4028}< \frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\Leftrightarrow A< B\)
XO
10 tháng 8 2020
Ta có \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)
\(=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4...2013.2015}{2.2.3.3...2014.2014}\right)\)
\(=-\left(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< \frac{-2014}{4028}< \frac{1}{2}=B\)
=> A < B
DP
20 tháng 8 2017
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
Bảo Online Math làm cho
$A=\frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{2014^2-1}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}=1-\frac{1}{2014}=\frac{2013}{2014}>-\frac{1}{2}$