八年级上数学分式计算题 较简单 求解答 谢谢！

Question

如题所述

Answer 1

原式=[1/(x^2y^3)]^(-1)*(x^2/y^3)^2

=(x^2y^3)(x^4/y^6)

=x^6/y^3

[1+1/(x-1)]/[x/(x^2-1)]=[x/(x-1)]/[x/(x^2-1)]=(x^2-1)/(x-1)=x+1.x=-2.则原式等于-1.

[x^2/(x-1)-2x/(x-1)]/[x/(x-1)]=[x(x-2)/(x-1)]/[x/(x-1)]=x-2=0

(x^2+y^2)/(x-y)-2xy/(x-y)=(x-y)^2/(x-y)=x-y


x/(x+2)=(x+2)/(x-2)

x(x-2)=(x+2)^2

x^2-2x=x^2+4x+4

6x+4=0

x=-2/3

[(3x-3)/(x^2-1)]/[3x/(x+1)]-1/(x-1)

={3(x-1)/[(x+1)(x-1)]}/[3x/(x+1)]-1/(x-1)

=[3/(x+1)]/[3x/(x+1)]-1/(x-1)

=1/x-1/(x-1)

=1/2-1/1

=-1/2

设甲打9000字需时x.则甲的速度为9000/x字/小时.

因乙打7200个字与甲打9000个需时一样.即乙的速度为7200/x字/小时.

由题意有

7200/x+9000/x=5400

(72+90)/x=54

x=162/54=3

则甲的速度为9000/3=3000字/小时

乙为7200/3=2400字/小时

Answer 2

1>x^6/y^3
2>原式=x+1 (x≠±1或x≠0) 代入x=-2，则原式=-1
3>原式=x-2 (x≠1或x≠0) 代入x=2，则原式=0
4>原式=x-y (x≠y)
5>x=-2/3
6>原式=-1/[x(x-1)] (x≠±1或x≠0) 代入x=2,则原式=-1/[2(2-1]=-1/2
7>设甲每小时打x个字，则乙每小时打7200x/9000=4/5x
x+4/5x=5400
9/5x=5400
x=3000
所以甲每小时打3000个字，乙每小时打2400个字追问

？

Answer 3

太幼稚了，耽误我时间追问

那就别来打扰我