Mấy chế zô giải giùm tui câu này:
Cho \(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{2013^2}-1\right)\left(\dfrac{1}{2014^2}-1\right)\) và \(B=-\dfrac{1}{2}.\)
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
27 tháng 1 2023
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
=1/x-1/x+2014
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
LB
20 tháng 4 2018
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
LB
20 tháng 4 2018
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
21 tháng 7 2018
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{2014}{x\left(x+2014\right)}\)
HS
21 tháng 7 2018
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+....+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2014}-\dfrac{x+2014}{x\left(x+2014\right)}-\dfrac{x}{x\left(x+2014\right)}\)
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}\)
\(=\dfrac{2014}{x\left(x+2014\right)}\)
6 tháng 5 2021
a)\(\left|\dfrac{1}{2}+x\right|-1=\dfrac{11}{2}\)
\(\Rightarrow\left|\dfrac{1}{2}+x\right|=\dfrac{11}{2}+1=\dfrac{13}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}+x=\dfrac{-13}{2}\\\dfrac{1}{2}+x=\dfrac{13}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=6\end{matrix}\right.\)
b)\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2014}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{2}{-3}.\dfrac{-3}{4}...\dfrac{2012}{-2013}.\dfrac{-2013}{2014}\)
\(=\dfrac{-1}{2014}\)
số nghịch đảo của 50% là:\(\dfrac{100}{50}=2\)
7 tháng 10 2017
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
7 tháng 10 2017
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
29 tháng 6 2021
Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right).......\cdot\left(\dfrac{1}{2014^2}-1\right)\)
\(=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)..........\left(\dfrac{1}{4056196}-1\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-4056195}{4056196}\)
\(=\dfrac{\left(-1\right).3}{2^2}.\dfrac{\left(-2\right).4}{3^2}...........\dfrac{\left(-2013\right).2015}{2014^2}\)
\(=\dfrac{\left(-1\right).\left(-2\right)..........\left(-2013\right)}{2.3......2014}.\dfrac{3.4......2015}{2.3.....2014}\)
\(=\dfrac{-1}{2014}.\dfrac{2015}{2}\)
\(=\dfrac{-2015}{4028}< \dfrac{-1}{2}\)
\(\Leftrightarrow A< B\)
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