=[(x 2)/(x-2)-1/(2-x)](x-4/x)
=(x 3)/(x-2).(x-4/x)
=(x^2-x-7)/(x^2-2x)
题目好像有问题，但思路是这样追问
为什么前面中括号里的多项式我和并得x+1/x-2
追答你是对的，我看错了～
求采纳
先化简,后求值:[(x^2-4)\/(x^2-4x+4)-x\/(x^2-2x)]*[x-(4\/x)]
=1-(x^2-2x)\/(x^2-4x+4)=1-x(x-2)\/(x-2)^2 =1-x\/(x-2) 这步可以采用直接变为（x-2-x）\/（x-2)=1-(x-2+2)\/(x-2) 看你自己哪种方法顺手了 =1-[1+2\/(x-2)]=2\/(2-x)
化简:【(x²-4)\/(x²-4x+4)-1\/(x-2)】*(x²-2x)\/(x+1)
解：[(x²-4)\/(x²-4x+4)-1\/(x-2)]*(x²-2x)\/(x+1)=[(x+2)(x-2)\/(x-2)²-1\/(x-2)]*x(x-2)\/(x+1)=[(x+2)\/(x-2)-1\/(x-2)]*x(x-2)\/(x+1)=[(x+1)\/(x-2)]*x(x-2)\/(x+1)=x ...
先化简,再求值,(x^2-1)\/(x^2-x-2)\/(x)\/(2x-4),其中x=1\/2
[(x²-1)\/(x²-x-2)]\/[x\/(2x-4)]=[(x²-1)\/(x²-x-2)][(2x-4)\/x]={[(x+1)(x-1)]\/[(x+1)\/(x-2)]}[2(x-2)\/x]=[(x-1)\/(x-2)][2(x-2)\/x]=[2(x-1)]\/x=[2(1\/2-1)]\/(1\/2)=1\/(1\/2)=2很高兴为您解答,祝你学习进步!【学习宝典】团队为您答题。
先化简,再求值:{[1\/(x^2-2x)-1\/(x^2-4x+4)]\/[1\/(x^2-2x)]},其中x=3
=1-(x^2-2x)\/(x^2-4x+4)=1-x(x-2)\/(x-2)^2 =1-x\/(x-2)这步可以采用直接变为（x-2-x）\/（x-2)=1-(x-2+2)\/(x-2)看你自己哪种方法顺手了 =1-[1+2\/(x-2)]=2\/(2-x)将x=3带入，解得原式=-2
一道数学化简题(先化简再求值) [1\/(x^2-2x)-1\/(x^2-4x+4)]\/2\/(x^...
1\/(x^2-4x+4)=1\/(x-2)^2 2\/(x^2-2x)=2\/x(x-2)分子分母乘以(x-2)[1\/x-1\/(x-2)]\/2\/x 分子分母乘以x [1-x\/(x-2)]\/2 分子通分 (x-2-x)\/(x-2)\/2 =1\/(2-x)(x-3)\/(2x-4)=[(x-3)\/(x-2)]\/2 =[1-1\/(x-2)]\/2 ps：当x=-1 1\/(2-x)=1\/3 ...
先化简,再求值:(x2\/x-2 - 4\/x-2)×1\/x2+2x.其中x=1\/4
（x2\/x-2 - 4\/x-2）×1\/x2+2x.= （x2-4)\/(x-2) ×1\/x2+2x.= (x+2)\/(x2+2x.)= 1 \/ x 其中x=1\/4 原式 = 4
...在求值:(X分之x2+4-4)除以x2+2x分之x2-4,其中x是方程x2+2x+1=10...
x²+2x+1=10 （x+1)²=10 x+1=3或x+1=-3 所以x=2或x=-4 【（x²+4)\/x-4】÷【（x²-4)\/(x²+2x)】=【（x²-4x+4)\/x】÷【(x+2)（x-2)\/x(x+2)】=【（x-2)²\/x】÷【（x-2)\/x】=【（x-2)²\/x】*【x\/（...
先化简 x²-4x+4\/x²-2x÷(x-4\/x)再从﹣根号5<x<根号5选取一个合适...
解 (x²-4x+4)\/(x²-2x)÷(x-4\/x)=(x-2)²\/x(x-2)÷[(x²-4)\/x]=[(x-2)\/x]×[x\/(x-2)(x+2)]=1\/(x+2)∵-√5<x<√5 ∴选x=0 ∴原式 =1\/(0+2)=1\/2
急!!两道计算题:先化简再求值:【x的平方减4】分之【x乘以(x-2...
x=2\/(2+√2）＝√2－2 [(x+2)^2]\/(x+2)-2(x-1),=(x+2)-2(x-1),=-x+3 =5-√2 6\/(x-1)-3\/(x-1)=3 3\/(x-1)=3 x-1=1 x=2
先化简,再求值:X^2-4X+4\/X^2+X\/(3\/X+1-X+1)+1\/X+2,其中X为方程X^2+...
(X^2-4X+4)\/(X^2+X)÷[3\/(X+1) -X+1] + 1\/(X+2)=(x-2)²\/[x(x+1)] ÷[(3-x²+1)\/(x+1)] + 1\/(X+2)=(x-2)²\/[x(x+1)] *(x+1)\/[-(x+2)(x-2)] + 1\/(X+2)=(x-2)\/[-x(x+2)] + 1\/(x+2)=(-x+2+x)\/[x(x+2)]...
