Question

a-b=根号2＋根号3b-c=根号2-根号3，求a方+b方+c方-ab-bc-ac的值

Answer 1

a-b=根号2＋根号3
b-c=根号2-根号3 相加
a-c=2根号2
a方+b方+c方-ab-bc-ac
=1/2(2a方+2b方+2c方-2ab-2bc-2ac)
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
=1/2[5+2根号6+5-2根号6+8]
=1/2*(18)
=9

Answer 2

解：a²+b²+c²-ab-bc-ac
=1/2(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2[(a-b)²+(b-c)²+(a-c)²]
带入a-b=√2＋√3
b-c=√2-√3
a-c=2√2

∴原式=
=1/2[（√2+√3）²+（√2-√3）²+（2√2）²]
=1/2（5+5+8）
=9