Question

已知a-b=2，b-c=1，求a*+b*+c*-ab-bc-ac的值 *=2

Answer 1

a-b=2，b-c=1
∴a-c=a-b+b-c=2+1=3
a²＋b²＋c²-ab-ac-bc
=½(a²-2ab+b²+a²-2ac+c²+b²-2bc+c²)
=½[﹙a-b)²+(a-c)²+(b-c)²]
=½[2²＋3²＋1²]
=½×14
=7

Answer 2

a*+b*+c*-ab-bc-ac=a(a-b-c)+b*+c*-bc=a(2-c)+b*+c*-bc=(2+b)(2-c)+b*+c*-bc=4+2(b-c)+(b-c)(b-c)=7