If x is real,then the sum of the maximum and | vedclass

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(D) Let $y = \frac{x^2+4x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2+4x+1$,which implies $(y-1)x^2 + (y-4)x + (y-1) = 0$.Since $x$ is real,the discriminant $

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(D) Let $y = \frac{x^2+4x+1}{x^2+x+1}$.Then $y(x^2+x+1) = x^2+4x+1$,which implies $(y-1)x^2 + (y-4)x + (y-1) = 0$.Since $x$ is real,the discriminant $D \ge 0$.$D = (y-4)^2 - 4(y-1)^2 \ge 0$.$(y-4)^2 - [2(y-1)]^2 \ge 0$.$(y-4-2y+2)(y-4+2y-2) \ge 0$.$(-y-2)(3y-6) \ge 0$.$(y+2)(3y-6) \le 0$.$(y+2)(y-2) \le 0$.Thus,$-2 \le y \le 2$.The minimum value is $-2$ and the maximum value is $2$.The sum of the maximum and minimum values is $2 + (-2) = 0$.

If $x$ is real,then the sum of the maximum and the minimum values of the expression $\frac{x^2+4x+1}{x^2+x+1}$ is

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