[Toán 8]Giải pt.

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trangbe132

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vanmanh2001

Chắc ý bạn là thế này
$\dfrac{1}{x^2 - x + 1} + 1 - \dfrac{x^2+2}{x^3 + 1} $
$= \dfrac{x+1}{x^3+1} + \dfrac{x^3+1}{x^3 + 1} - \dfrac{x^2 + 2}{x^3 + 1}$
$= \dfrac{x+1 + x^3 + 1 - x^2 - 2}{x^3+1}$
$= \dfrac{x^3 - x^2 + x}{x^3 + 1}$
$= \dfrac{x(x^2 - x + 1)}{(x+1)(x^2 - x + 1)}$
$= \dfrac{x}{x + 1} = \dfrac{99}{100}$
 
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