a) cho A = (1/2^2-1) (1/3^2-1) .... (1/2013^2-1) (1/2014^2-1)và B= -1/2 . So sánh A và B
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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
NH
2
16 tháng 8 2015
$A=\frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{2014^2-1}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}=1-\frac{1}{2014}=\frac{2013}{2014}>-\frac{1}{2}$
QK
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 \(10A>10B\Rightarrow A>B\)
BB
0
HV
1
17 tháng 2 2020
\(\Rightarrow A-B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4026}\)
\(B>1+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{4026}=\frac{1}{2}+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{4026}\right)=\frac{1}{2}+\left(A-B\right)\)
\(\Rightarrow B>\frac{1}{2}+\left(A-B\right)\left(1\right)\)
\(A-B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4026}< \frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}=\frac{2013}{2}\)
\(\Rightarrow A-B< \frac{2013}{2}\Rightarrow\frac{A-B}{2013}< \frac{1}{2}\left(2\right)\)
Cộng (1) với (2)
\(\Rightarrow\frac{A-B}{2013}+\frac{1}{2}+\left(A-B\right)< \frac{1}{2}+B\Rightarrow\frac{A-B}{2013}+\left(A-B\right)< B\Rightarrow\frac{2014\left(A-B\right)}{2013}< B\Rightarrow\frac{A-B}{B}< \frac{2013}{2014}\)
\(\Rightarrow\frac{A-B}{B}+1< \frac{2013}{2014}+1\Rightarrow\frac{A}{B}< 1\frac{2013}{2014}\left(đpcm\right)\)
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
Ngu ngờ ngáo !