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\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)
\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)
\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)
\(=\frac{6}{2011}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)\cdot\cdot\cdot\cdot\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)
\(A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)\left(\frac{-15}{16}\right)\cdot\cdot\cdot\left(\frac{-4052168}{4052169}\right)\left(\frac{-4056195}{4056196}\right)\)
\(A=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot....\cdot\frac{-2012\cdot2014}{2013\cdot2013}\cdot\frac{-2013\cdot2015}{2014\cdot2014}\)
\(A=\frac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot....\cdot\left(-2012\right)\cdot\left(-2013\right)}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\cdot\frac{3\cdot4\cdot5\cdot....\cdot2014\cdot2015}{2\cdot3\cdot4\cdot....\cdot2013\cdot2014}\)
\(A=\frac{-1}{2014}\cdot\frac{2015}{2}=\frac{-2015}{4028}\)
Ta thấy \(\frac{-2015}{4028}< \frac{-1}{2}\) \(\Rightarrow A< B\)
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
Vì \(\frac{1}{2^2}>0\)
............
\(\frac{1}{2014^2}>0\)
=> A = \(\left(\frac{1}{2^2}\right)\left(\frac{1}{3^2}\right)...\left(\frac{1}{2014^2}\right)>0\)
B = \(-\frac{1}{2}<0\)
Vậy A > B
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