已知a与b互为相反数，c与d互为倒数，m是绝对值为4的负数，求23a+23b+(cd)2011−32m的值．

Question

已知a与b互为相反数，c与d互为倒数，m是绝对值为4的负数，求(2)/(3)a+(2)/(3)b+(cd)2011−(3)/(2)m的值．

Answer 1

由题意得：a+b=0，cd=1，m=-4
原式=(2)/(3)(a+b)+(cd)2011−(3)/(2)m=(2)/(3)×0+12011−(3)/(2)×(−4)=0+1+6=7