Question

3、已知x分之1+y分之1=5,求x-2xy+y分之2x+3xy+2y的值

4、先化简，再求值[（x²-1分之3x＋4）-（x-1分之2）]÷x²-2x＋1分之x＋2 ，其中x是不等式组{x+4＞0，2x+5＜1的整数解

Answer 1

解
1/x+1/y=5
即(y+x)/xy=5
∴y+x=5xy
∴(2x+3xy+2y)/(x-2xy+y)
=[2(x+y)+3xy]/[(x+y)-2xy]
=[2×5xy+3xy]/(5xy-2xy)
=(13xy)/(3xy)
=13/3

x+4>0
x>-4
2x+5<1
x<-2
∴-2<x<-4
∵x为整数
∴x=-3
[(3x+4)/(x²-1)-2/(x-1)]÷(x+2)/(x²-2x+1)
=[(3x+4)/(x-1)(x+1)-2(x+1)/(x-1)(x+1)]×(x-1)²/(x+2)
=[(3x+4-2x-2)/(x-1)(x+1)}×(x-1)²/(x+2)
=(x+2)/(x-1)(x+1)×(x-1)²/(x+2)
=(x-1)/(x+1)
=(-3-1)/(-3+1)
=-4/(-2)
=2

Answer 2

你是 初二的 吧 让我想想

追问

我已经知道喽