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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 \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
![](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 \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
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\(a.\frac{10^{11}-1}{10^{12}-1}<\frac{10^{10}+1}{10^{10}+1}\)
\(b.\)(\(\frac{1}{80}\))\(^7\)\(>\)(\(\frac{1}{243}\))\(^6\)
Tick mình nha
a) Ta thấy B < 1 và 9 > 0 nên ta có: \(B< \frac{10^{2002}+1+9}{10^{2003}+1+9}\)
\(< \frac{10^{2002}+10}{10^{2003}+10}\)
\(< \frac{10\left(10^{2001}+1\right)}{10\left(10^{2002}+1\right)}\)
\(< \frac{10^{2001}+1}{10^{2002}+1}=A\)
Vậy \(A>B\)