1,已知a^2+2a+b^2-6b+10=0,求a^b的值
a^2 + 2a + b^2 - 6b + 10
= (a^2 + 2a + 1) + (b^2 - 2*3*b + 9)
= (a + 1)^2 + (b - 3)^2
= 0
a+1=0
b-3=0
a = -1
b = 3
a^b=(-1)^3=-1
2,已知a-b=-2,b-c=3.求a^2+b^2+c^2-ab-bc-ac的值
因为a-b=-2,b-c=3,所以a-c=a-b+b-c=1
a^2+b^2+c^2-ab-bc-ac
=0.5*2(a^2+b^2+c^2-ab-bc-ac)
=0.5*(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=0.5*(a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2)
=0.5*((a-b)^2+(a-c)^2+(b-c)^2)
=0.5*((-2)^2+1^2+3^2)
=7