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)

题目

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

答案

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