If x is real,then the minimum value of y = fr | vedclass

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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.Rearranging the terms as a quadratic in $x$:

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$\frac{1}{3}$

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(B) Let $y = \frac{x^2-x+1}{x^2+x+1}$.Multiplying both sides by $(x^2+x+1)$,we get $y(x^2+x+1) = x^2-x+1$.Rearranging the terms as a quadratic in $x$:$(y-1)x^2 + (y+1)x + (y-1) = 0$.Since $x$ is real,the discriminant $D \geq 0$.$D = (y+1)^2 - 4(y-1)^2 \geq 0$.$(y+1)^2 - [2(y-1)]^2 \geq 0$.Using $a^2-b^2 = (a-b)(a+b)$:$[(y+1) - 2(y-1)][(y+1) + 2(y-1)] \geq 0$.$(-y+3)(3y-1) \geq 0$.$(y-3)(3y-1) \leq 0$.This inequality holds for $\frac{1}{3} \leq y \leq 3$.Thus,the minimum value of $y$ is $\frac{1}{3}$.

If $x$ is real,then the minimum value of $y = \frac{x^2-x+1}{x^2+x+1}$ is

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