\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{36}}\)
Vì \(2^{35}\frac{1}{2^{36}}\)
Vậy \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
(1/32)^7 = (1/2^5)^7 =(1/2)^35 > ( 1/2 ) ^36 = (1/2^4 )^9 = ( 1/ 16 ) ^9