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Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{18}{19}.\frac{19}{20}\)
\(=\frac{1.2....18.19}{2.3...19.20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Vậy A > 1/21
Bài làm
Ta có: \(\left(-\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{4}\right)^2\right]^5=\left(\frac{1}{4}\right)^{10}\)
Mà \(2< 10\)
=> \(\left(\frac{1}{4}\right)^2< \left(\frac{1}{4}\right)^{10}\)
Hay \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)
Vậy \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)
# Học tốt #
\(\left(\frac{-1}{64}\right)^5=\left(\left(\frac{-1}{4}\right)^3\right)^5=\left(\frac{-1}{4}\right)^{15}\)
\(\left(\frac{-1}{4}\right)^{15}< \left(\frac{-1}{4}\right)^7\Leftrightarrow\left(\frac{-1}{64}\right)^5< \left(\frac{-1}{4}\right)^7\)
\(\left(\frac{-1}{64}\right)^5=-\frac{1}{64^5}=-\frac{1}{\left(4^3\right)^5}=-\frac{1}{4^{15}}\)
\(\left(-\frac{1}{4}\right)^7=-\frac{1}{4^7}\)
\(-\frac{1}{4^{15}}>-\frac{1}{4^7}\)
\(\Rightarrow\left(-\frac{1}{64}\right)^5>\left(-\frac{1}{4}\right)^7\)
Ta có: \(\left(\frac{-1}{4}\right)^{40}=\left[\left(\frac{-1}{4}\right)^2\right]^{20}=\left(\frac{1}{16}\right)^{20}\)
\(\left(\frac{-1}{5}\right)^{34}=\left[\left(\frac{-1}{5}\right)^2\right]^{17}=\left(\frac{1}{25}\right)^{17}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{20}>\left(\frac{1}{25}\right)^{17}\)
Vậy \(\left(\frac{-1}{4}\right)^{40}>\left(\frac{-1}{5}\right)^{34}\)