Question

已知a-b=根号3+根号2,b-c=根号3-根号2.求a²+b²+c²-ab-bc-ca的值

Answer 1

a-b=√3+√2 1

b-c=√3-√2 2

1式+2式得

a-c=2√3

a²+b²+c²-ab-bc-ca

=1/2*(2a^2+2b^2+2c^2-2ab-2bc-2ca)

=1/2(a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2)

=1/2[(a-b)^2+(a-c)^2+(b-c)^2]

=1/2[(√3+√2)^2+(2√3)^2+(√3-√2)^2]

=1/2(3+2√6+2+12+3-2√6+2)

=1/2*22

=11