Question

已知a与b互为相反数,c与d互为倒数,x=4(a-2)-(a-3b),y=c2d(cd+1/c-1),求4分之x+y-8分支3x-5y

Answer 1

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

追问

^是什么意思

追答

次方的意思，c^2表示c的2次方

Answer 2

由题知a+b=0，cd=1

x=4a-8-a+3b=3a+3b-8=-8
y=c(1+1/c-1)=1
所以x/4+y-3x/8-5y=-2+1+3-5=-3