题目

已知a与b互为相反数,c与d互为倒数,x=4(a-2)-(a-3b),y=c^2d+d^2-(d/c+c-1),求(x+y)/4-(3x-5y)/8的值

提问时间：2020-07-31

答案

a与b互为相反数,c与d互为倒数
a+b=0
cd=1
x=4(a-2)-(a-3b)
=3a+3b-8
=3*(a+b)-8=-8
d=1/c
y=c^2d+d^2-(d/c+c-1)
=c+1/c^2-1/c^2-c+1
=1
(x+y)/4-(3x-5y)/8
=(-8+1)/4-(-24-5)/8
=-7/4+29/8
=15/8