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(A) Given the quadratic equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$.Let the coefficients be $A = (b + c - 2a)$,$B = (c + a - 2b)$,an

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(A) Given the quadratic equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$.Let the coefficients be $A = (b + c - 2a)$,$B = (c + a - 2b)$,and $C = (a + b - 2c)$.Sum of the coefficients is $A + B + C = (b + c - 2a) + (c + a - 2b) + (a + b - 2c) = (a + a - 2a) + (b + b - 2b) + (c + c - 2c) = 0$.Since the sum of the coefficients is $0$,one root of the quadratic equation must be $x = 1$.Since $a, b, c \in \mathbb{Q}$,the coefficients $A, B, C$ are rational numbers.For a quadratic equation with rational coefficients,if one root is rational,the other root must also be rational (as the product of roots $C/A$ is rational).Therefore,the roots are rational.

If $a, b, c \in \mathbb{Q}$,then the roots of the equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$ are

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