已知a&b互为相反数,c&d互为倒数,x=3(a-1)-(a-2b),y=c+d^2-[(d/c)+c-1],求[(2x+y)/3]-[(3x-y)/6]的值
问题描述:
已知a&b互为相反数,c&d互为倒数,x=3(a-1)-(a-2b),y=c+d^2-[(d/c)+c-1],求[(2x+y)/3]-[(3x-y)/6]的值
答
把b=-a,c=1/d,带入得到x=3(a-1)-(a-2b)=-3,y=c+d^2-[(d/c)+c-1]=1;再代入表达式得到(-5/3)-(-10/6)=0