Ta có: \(\left(\dfrac{1}{32}\right)^7=\dfrac{1^7}{32^7}=\dfrac{1}{2^{35}}\\ \left(\dfrac{1}{16}\right)^3=\dfrac{1^3}{16^3}=\dfrac{1}{2^{12}}\)
Do 2^35 > 2 ^ 12 suy ra \(\dfrac{1}{2^{35}}< \dfrac{1}{2^{12}}\Rightarrow\left(\dfrac{1}{32}\right)^7< \left(\dfrac{1}{16}\right)^4\)