Question

已知a-b=2+根号3,b-c=2-根号3,则a平方+b平方+c平方-ab-ac-bc的值为多少

Answer 1

a-b=2+根号3,b-c=2-根号3

相加

a-c=4

原式= (2a²+2b²+2c²-2ab-2bc-2ac)/2

=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2

=[(a-b)²+(b-c)²+(a-c)²]/2

=(7+4√3+7-4√3+16)/2

=15