分式2道题⒈已知1/x+1/y=5,则(2x-3xy+2y)/(x+2xy+y)=?⒉分式〔2(x-1)〕/(3x-4)的值为正值时,则x的取值范围是?

Question

分式2道题
⒈已知1/x+1/y=5,则(2x-3xy+2y)/(x+2xy+y)=?
⒉分式〔2(x-1)〕/(3x-4)的值为正值时,则x的取值范围是?

Answer 1

⒈已知1/x+1/y=5,则(2x-3xy+2y)/(x+2xy+y)=?
(x+y)/(xy)=5
x+y=5xy
原式=[2(x+y)-3xy]/[(x+y)+2xy]
=(10xy-3xy)/(5xy+2xy)
=1
⒉分式〔2(x-1)〕/(3x-4)的值为正值时,则x的取值范围是?
2(x-1)/(3x-4)>0
上式可以有如下二种:
(1)x-1>0且3x-4>0
解得:x>4/3
(2)x-1