Question

已知a-b=5,b-c=3,求a²+b²+c²-ab-bc-ac的值.

Answer 1

a-b=5,b-c=3
两式相加得a-c=8
2(a²+b²+c²-ab-bc-ac)
=2a²+2b²+2c²-2ab-2bc-2ac
=(a-b)²+(b-c)²+(a-c)²
=25+9+64=98
a²+b²+c²-ab-bc-ac=49
祝你好运

Answer 2

a-b=5,b-c=3
a-c=8
a²+b²+c²-ab-bc-ac
=1\2[(a-b)^2+(b-c)^2+(a-c)^2]
=49

Answer 3

a-b=5,b-c=3
a-c=8
a²+b²+c²-ab-bc-ac
=a²-2ab+b²+c²+ab-bc-ac
=(a-b)²+(ab-bc)+(c²-ac)
=(a-b)²+b(a-c)+c(c-a)
=(a-b)²+(a-c)(b-c)
=5²+8*3
=49