八年级上数学分式计算题 较简单 求解答 谢谢!
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
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
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那就别来打扰我