设X1,X2是方程X^2-2aX+a+b=0两实数根,求(X1-1)^2+(X2-1)^2的最小值

X1,X2是方程X^2-2aX+a+b=0两实数根
x1+x2=2a x1*x2=a+b
且△=(-2a)^2-4(a+b)≥0a^2≥a+b=x1*x2
(X1-1)^2+(X2-1)^2
=(x1^2-2x1+1)+(x2^2-2x2+1)
=(x1+x2)^2-2x1*x2-2(x1+x2)+2
=4a^2-4a+2-2x1*x2
≥4a^2-4a+2-2a^2
=2(a^2-2a+1)
=2(a-1)^2
≥0
所以最小值为0
希望能帮到你,祝学习进步O(∩_∩)O