\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7>\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{243}\right)^6\)
\(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1^7}{\left(3^4\right)^7}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1^6}{\left(3^5\right)^6}=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\) nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
ta có
=243^6
=(3^5)^6=3^30>3^28
=(3^4)^7
=81^7>80^...
=>(1/80)^7>(1/243)^6