Question

已知a与b互为相反数,c与d互为倒数,x=3(3-1)-(a-2b),y=c^2d+d^2-(d/c+c-1),求2x+y/3 - 3x-2y/6的指

已知a与b互为相反数,c与d互为倒数,x=3(a-1)-(a-2b),y=c^2d+d^2-(d/c+c-1),求2x+y/3 - 3x-2y/6的值

Answer 1

x=3(a-1)-(a-2b)
=3a-3-（a+2a)
=-3
y=c^2d+d^2-(d/c+c-1)
=c+d^2- (d^2+1/d-1)
=1/d+d^2-d^2-1/d+1
=1

2x+y/3 - 3x-2y/6
=-x
=3

Answer 2

a+b=0
cd=1
x=3a-3-a+2b
=2a+2b-3
=0-3=-3
y=c(cd)+d²-d/(1/d)-c+1
=c+d²-d²-c+1
=1

则原式=(4x+2y-3x+2y)/6
=(x+4y)/6
=(-3+4)/6
=1/6