a-b=1+根号2,b-c=1-根号2
a-c=2
所以
a²+b²+c²-ab-bc-ac
=1/2(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2[(a-b)²+(b-c)²＋(c-a)²]
=1/2[(1+根号2)²+(1-根号2)²＋(-2)²]
=1/2[3＋2根号2＋3-2根号2＋4]
=1/2×10
=5