Ta có:

a) $(−0,1)2 . (−0,1)^{4} = (−0,1)^{2+4} = (−0,1)^{6} = (−0,1)^{3.2} = [(−0,1)^{3}]^{2}$

=>$(−0,1)2 . (−0,1)^{4}$ = $[(−0,1)^{3}]^{2}$;

b) $(\frac{1}{2})^{8} : (\frac{1}{2})^{2} = (\frac{1}{2})^{8−2} = (\frac{1}{2})^{6} = (\frac{1}{2})^{3+3} = (\frac{1}{2})^{3}. (\frac{1}{2})^{3}$

=>$(\frac{1}{2})^{8} : (\frac{1}{2})^{2}$ = $(\frac{1}{2})^{3} . (\frac{1}{2})^{3}$;

c) $9^{8} : 27^{3}= (3^{2})^{8} : (3^{3})^{3} = 3^{2.8} : 3^{3.3} = 3^{16} : 3^{9} = 3^{7} = 3^{2+5} = 3^{2} . 3^{5}$

=>$9^{8} : 27^{3}$ = $ 3^{2}  . 3^{5}$ ;

d) $(\frac{1}{4})^{7} . 0,25= (\frac{1}{4})^{7} .\frac{1}{4} = (\frac{1}{4})^{7+1} = (\frac{1}{4})^{8} = (\frac{1}{4})^{2.4} = [(\frac{1}{4})^{2}]^{4}$

=>$(\frac{1}{4})^{7} . 0,25$ = $[(\frac{1}{4})^{2}]^{4}$

e) $ [(−0,7)^{2}]^{3} = [(0,7)^{2}]^{3} =  (0,7)^{2.3} =[(−0,7)^{3}]^{2}$

=>$[(−0,7)^{2}]^{3}$ = $[(−0,7)^{3}]^{2}$