(x-1分之x + x-1分之1)除以x^2-2x+1分之x+1其中x=-2

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(x-1分之x+x-1分之1)除以x^2-2x+1分之x+1其中x=-2(x-1分之x+x-1分之1)除以x^2-2x+1分之x+1其中x=-2(x-1分之x+x-1分之1)除以x^2-2x+1分之x

(x-1分之x + x-1分之1)除以x^2-2x+1分之x+1其中x=-2
(x-1分之x + x-1分之1)除以x^2-2x+1分之x+1
其中x=-2

(x-1分之x + x-1分之1)除以x^2-2x+1分之x+1其中x=-2
答:
[x/(x-1)+1/(x-1)]÷[(x+1)/(x²-2x+1)]
=[(x+1)/(x-1)]÷[(x+1)/(x-1)²]
=[(x+1)/(x-1)]×[(x-1)²/(x+1)]
=x-1

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

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