2 1( )=  −  = − =15 16 4 43  2B, TA CÓ

3 .27 : 3 7 : 2 9 7 : 2 1

( )

=  −  = − =

15 16 4 4

3

  2

b, Ta có: 24 : 3 ^{4 4} − 32 :16 ^{12 12} = ( ^{24 : 3} ) ( ^{4 − 32 :16} ) ^{12 = 8 4 − 2 12 = 2 12 − 2 12 = 0}