若a与2b互为负倒数,-c与d/2互为相反数,m与n互为相反数,求2ab-2c+d+m÷n

问题描述:

若a与2b互为负倒数,-c与d/2互为相反数,m与n互为相反数,求2ab-2c+d+m÷n

你好!
由题意 a*2b = -1 ,-c+ d/2 = 0 即 -2c+d = 0 ,m÷n= -1
2ab-2c+d+m÷n = - 1 + 0 - 1 = - 2

  由题可得,a=-1/(2b),-c=-d/2,m=-n
  
  ∴2ab=-[1/(2b)]*2b=-1,-2c+d=-d+d=0,m/n=-n/n=-1
  
  ∴2ab-2c+d+m÷n=-1-1=-2

∵a与2b互为负倒数
∴2ab=-1
∵-c与d/2互为相反数
∴-c+d/2=0
∴2(-c+d/2)=0
即 -2c+d=0
∵m与n互为相反数
∴m÷n=-1
所以 2ab-2c+d+m÷n=2ab+(-2c+d)+m÷n =-1+0+(-1)=-2

=2x1/2bxb-2xd/2-1
=1-d-1
=-d

a与2b互为负倒数,则a×2b=2ab=-1
-c与d/2互为相反数,则 -c+d/2=(-2c+d)/2=0,-2c+d=0
m与n互为相反数,则m=-n,m÷n=-1
2ab-2c+d+m÷n
=-1+0+(-1)
=-2