Question

已知：a=-2011,b=2010,c=-2009,求a²+b²+c²+ab+bc-ac

Answer 1

分析：先求2(a²+b²+c²+ab+bc-ac),然后配方

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

=(a²+2ab+b²)+(a²-2ac+c²)+(b²+2bc+c²)

=(a+b)²+(a-b)²+(b+c)²

=(-2011+2010)²+(-2011+2009)²+(2010-2009)²

=1+(-2)²+1=4

故a²+b²+c²+ab+bc-ac=4/2=2