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\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1^7}{\left(3^4\right)^7}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1^6}{\left(3^5\right)^6}=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\) nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
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Ta có:\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{79}{80}\Rightarrow A< \frac{2}{3}\cdot\frac{4}{5}\cdot...\cdot\frac{80}{81}\)
\(\Leftrightarrow A^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{79}{80}\cdot\frac{80}{81}=\frac{1}{81}=\left(\frac{1}{9}\right)^2\)
\(\Leftrightarrow A< \frac{1}{9}\)
Nhớ tk mk nha!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7>\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{243}\right)^6\)
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1^7}{\left(3^4\right)^7}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1^6}{\left(3^5\right)^6}=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\) nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1}{5}.4=\frac{4}{5}\)
\(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}< \frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{1}{10}.4=\frac{2}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}< \frac{1}{15}+\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{1}{15}.4=\frac{4}{15}\)
\(\frac{1}{17}+\frac{1}{18}< \frac{1}{17}.2=\frac{2}{17}\)
\(\Rightarrow B< \frac{4}{5}+\frac{2}{5}+\frac{4}{15}+\frac{2}{17}=\frac{404}{255}=1\frac{149}{255}< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\frac{10^{11}-1}{10^{12}-1}<\frac{10^{10}+1}{10^{10}+1}\)
\(b.\)(\(\frac{1}{80}\))\(^7\)\(>\)(\(\frac{1}{243}\))\(^6\)
Tick mình nha
Ta có :
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)7=\frac{1}{3^{28}}\)
\(\left(\frac{1}{213}\right)^6>\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Có : (1/80)^7 < (1/64)^7 = [(1/2)^6]^7 = (1/2)^42
(1/213)^6 > (1/218)^6 = [(1/2)^7]^6 = (1/2)^42
=> (1/80)^7 < (1/213)^6
Tk mk nha