Question

先化简代数式再求值(3/x+1 - x+1)÷x²-2x/x+1，其中x满足方程x/x-1+1/x=1

Answer 1

你好！

[ 3/(x+1) - x+1 ] ÷ (x² - 2x)/(x+1)
= [ 3 - (x-1)(x+1) ] / (x+1) * (x+1)/(x² - 2x)
= (4 - x²) / (x² - 2x)
= (2+x)(2-x) / [x(x-2)]
= - (2+x)/x

x/(x-1) + 1/x = 1
x² + x-1 = x(x-1)
x = 1/2

代入得 原式 = - 5

追问

我错了 我以为你只做了一步 他做了两步 你看嘛 我赞都没赞他

追答

考试的时候不用写他那么多，只要写出关键步骤就可以了

追问

喔

Answer 2

=[3/(x+1)-(x-1)]÷(x²-2x)/(x+1)
=[3/(x+1)-(x-1)(x+1)/(x+1)]÷x(x-2)/(x+1)
=[3-(x+1)(x-1)]/(x+1)÷x(x-2)/(x+1)
=[3-x²+1)]/(x+1)÷x(x-2)/(x+1)
=(4-x²)/(x+1)/(x+1)÷x(x-2)/(x+1)
=(2+x)(2-x)/(x+1)x(x-2)/(x+1)
=-(x+2)(x-2)/(x+1)*(x+1)/x(x-2)
=-(x+2)/x

x/(x-1) + 1/x = 1
(x² + x-1) / x(x-1)=1
x² + x-1 = x(x-1)
x² + x-1 = x² -x
2x=1
x = 1/2

(3/x+1 - x+1)÷x²-2x/x+1
=-(x+2)/x
=-(1/2+2)÷1/2
=-5/2*2
=-5