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

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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

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
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

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
a与b互为相反数
所以:a+b=0
c与d互为倒数
所以:cd=1
x=3(a-1)-(a-2b),
=2a+2b-3
=2(a+b)-3
=-3
cd=1
两边同时除以c
得出:d/c=1/c
y=c^2d+d^2-(d/c+c-1)
=cd*c+d^2-(1/c+c-1)
=d^2-1/c+1
=(cd^2-1+c)/c
=1
所以:(2x+y/3)-(3x-2y/6)
=3+2/3
=11/3