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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2-x+1$,which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.Since $x$ is real,the discriminant $D

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$2$

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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2-x+1$,which implies $(y-1)x^2 + (y+1)x + (y-1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (y+1)^2 - 4(y-1)^2 \ge 0$.$(y+1)^2 - [2(y-1)]^2 \ge 0$.$(y+1-2y+2)(y+1+2y-2) \ge 0$.$(3-y)(3y-1) \ge 0$.$(y-3)(3y-1) \le 0$.Thus,$\frac{1}{3} \le y \le 3$.The greatest value is $3$ and the least value is $\frac{1}{3}$.The difference is $3 - \frac{1}{3} = \frac{9-1}{3} = \frac{8}{3}$.

If $x$ is real,then the difference between the greatest and least values of $\frac{x^2-x+1}{x^2+x+1}$ is

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