分式2道题
来源:百度知道 编辑:UC知道 时间:2024/09/28 08:44:47
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?
⒉分式〔2(x-1)〕/(3x-4)的值为正值时,则x的取值范围是?
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<0 且3x-4<0
解得:x<1
综上所述,x<1或x>4/3
1. (x+y)/xy = 5
原式 = (10xy-3xy)/(5xy+2xy) = 1
2. x < 3/4 或 x > 1
1.(2x-3xy+2y)/(x+2xy+y)=[2/x+2/y-3]/[1/x+1/y+2]=10-3/5+2=1
2.(x-1)*(3x-4)大于0,故x大于4/3或x小于1
(x+y)/xy = 5
原式 = (10xy-3xy)/(5xy+2xy)=1
x< 3/4 或 x > 1