已知AB为相反数 CD为倒数 X=8(a-1)-3(a-5/3b) Y=c²d+d²-(d/c+c-

即a+b=0
cd=1
所以x=8a-8-3a+5b
=5(a+b)-8
=0-8
=-8
y=c(cd)-d²+d/c-c+2
=(c-c)+(-cd²+d)/c-2
=(-d+d)/c-2
=-2
所以原式=(4x-2y-3x-2y)/6
=(x-4y)/6
=(-8+8)/6
=0
再问： c(cd)-d²+d/c-c+2 Ϊʲô��-d²
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