Question

已知a-b=√3+√2,b-c=√3-√2,求a2+b2+c2-ab-ac-bc的值

急求啊~~~

Answer 1

a-b=√3+√2,b-c=√3-√2
相加
a-c=2√3

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+(a-c)²]/2
=(5+2√6+5-2√6+12)/2
=11

Answer 2

a-b=√3+√2
b-c=√3-√2
两式相加得：a-c=2√3
a^2+b^2+c^2-ab-ac-bc
=1/2{(a-b)^2+(b-c)^2+(a-c)^2}
=1/2{(√3+√2)^2+(√3-√2)^2+(2√3)^2}
=1/2{3+2+3+2+12}
=11