b) Ta có: \(9^5=3^{10}\) ; \(27^3=3^9\)
Mà \(3^{10}>3^9\) => \(9^5>27^3\) 9 (đpcm)
c) Ta có: \(\left(\dfrac{1}{8}\right)^6=\dfrac{1}{2^{18}}\) ; \(\left(\dfrac{1}{32}\right)^4=\dfrac{1}{2^{20}}\)
Mà \(\dfrac{1}{2^{18}}>\dfrac{1}{2^{20}}\) => \(\left(\dfrac{1}{8}\right)^6>\left(\dfrac{1}{32}\right)^4\) (đpcm)