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-2),求(2x+y/3)-(3x+2y/6)的值
问题描述：
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-2),求(2x+y/3)-(3x+2y/6)的值
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答
a与b互为相反数,c与d互为倒数所以a+b=0,cd=1x=3a-3-a+2b=2a+2b-3=2(a+b)-3=0-3=-3y=c(cd)+d²-d/c-c+2=c+d²-d/c-c+2=d(d-1/c)+2=d(cd-1)/c+2=d(1-1)/c+2=2所以原式=(4x+2y-3x-2y)/6=x/6=-3/6=-1/2...