Question

a=m+20,b=m+19,c=m+21,则代数式a的平方+b的平方+c的平方-ab-bc-ca 的值?

Answer 1

即a-n=1

b-c=-2

c-a=1

所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2

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

=[(a-b)²+(b-c)²+(c-a)²]/2

=(1+4+1)/2

=3