Let a, b, c be rational numbers and satisfy (a+b+c-6)2+(2a+3b+c-11)2+(3a-b-c+2)2=0.

Let a, b, c be rational numbers and satisfy (a+b+c-6)2+(2a+3b+c-11)2+(3a-b-c+2)2=0.

By title
A+b+c-6=0
2A+3b+c-11=0
3A-b-c+2=0
I.e.
A+b+c=6
2A+3b+c=11
3A-b-c=-2
The first formula plus the third formula
4A=4
Therefore a=1
Substitute into the first expression
B+c=5
Substitute into the second expression
3B+c =9
These two expressions are subtracted
2B =4
So b=2
C=3
2A+b-c=1

If a is a rational number, then the value () of the entire equation a2(a2-2)-2a2+4 A. Not negative B. Constant is positive C. Constant negative D. Not equal to zero

A2(a2-2)-2a2+4=a4-2a2-2a2+4=a4-4a2+4=(a2-2)2≥0,
A is a rational number, a2 is not equal to 2, integer=0
The value of a2(a2-2)-2a2+4 is a positive number.
Therefore, B.