a, $111111_{2}$ = ($2^{5}$ + $2^{4}$ + $2^{3}$ + $2^{2}$ + 2 + 1)$_{10}$ = 63$_{10}$
b,4B$_{16}$ = (4*$16^{1}$ + 11*16$^{0}$ )$_{10}$ = 75$_{10}$
c, $1001111_{2}$ = ($2^{6}$ + $2^{3}$ + $2^{2}$ + $2^{1}$ + 1)$_{10}$= 79$_{10}$
= (4*$16^{1}$ + 15*16$^{0}$ )$_{10}$ = 4F$_{16}$
#xin CTLHHN, chúc bạn học tốt