Rút gọn $\left[\dfrac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right]\div \dfrac{x^2+x}{x^3+x}$

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1. Đáp án:

   `(x^2+1)/(x+1)` 

   Giải thích các bước giải:

   `( (x-1)^2/(3x+(x-1)^2) – (1-2x^2+4x)/(x^3-1) + 1/(x-1) ) : (x^2+x)/(x^3+x)`

    `=( (x-1)^2/(x^2+x+1) – (1-2x^2+4x)/(x^3-1) + 1/(x-1) ) : (x(x+1))/(x(x^2+1))`

   `= ((x-1)(x-1)^2-(1-2x^2+4x)+x^2+x+1)/(x^3-1) : (x+1)/(x^2+1)`

   `= (x^3-1)/(x^3-1) : (x+1)/(x^2+1) = 1 : (x+1)/(x^2+1)`

   `= 1 . (x^2+1)/(x+1) = (x^2+1)/(x+1)`

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2. `[(x-1)^2/(3x+(x-1)^2)-(1-2x^2+4x)/(x^3-1)+1/(x-1)]:(x^2+x)/(x^3+x)`

   `=[(x-1)^2/(x^2+x+1)-(1-2x^2+4x)/(x^3-1)+1/(x-1)]:(x+1)/(x^2+1)`

   `=[(x-1)^2/(x^2+x+1)-(1-2x^2+4x)/(x^3-1)+1/(x-1)]:(x+1)/(x^2+1)`

   `=[((x-1)^3-(1-2x^2+4x)+(x^2+x+1))/(x^3-1)]:(x+1)/(x^2+1)`

   `=[(x^3-1)/(x^3-1)].(x^2+1)/(x+1)`

   `=(x^2+1)/(x+1)`

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