Question

已知a-b=5,b-c=3求a的平方+b的平方+c的平方-ab-bc-ac的值

Answer 1

a-b=5,b-c=3
∴a-c=a-b+b-c=5+3=8
a²+b²+c²-ab-bc-ac
=½(2a½+2b½+2c½-2ab-2ac-2bc)
=1/2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2×﹙5²+8²+3²)
=1/2×98
=49

Answer 2

a-b=5
b-c=3
两式子相加得a-c=8
a^2+b^2+c^2-ab-bc-ac
=a(a-b)+b(b-c)-c(a-c)
=5a+3b-8c
=5(a-c)+3(b-c)
=5×8+3×3
=49