Question

谁帮我解答下这两道初一数学题 -----给下过程

1. a²+b²-4a+8b+20=0 求a+b的值

2.a=2005，b=2006，c=2007.求a²+b²+c²-ab-bc-ca的值

Answer 1

1、a²+b²-4a+8b+20

= a²-4a+4+b²+8b+16

= (a-2)^2 + (b+4)^2

=0

所以 a= 2 b= -4

所以 a+b = -2

2、 a²+b²+c²-ab-bc-ca

乘以2后再除以2 = 0.5(a²-2ab+b²+a²-2ac+c²+b²-2bc+c²)

=0.5[(a-b)^2+(a-c)^2+(b-c)^2]

=0.5[1+4+1]

=3

Answer 2

a2+4a+4+b2+8b+16=0
（a-2）2+（b+4）2=0
a=2 b=-4
a+b=-2

a2+b2+c2-ab-bc-ac
=(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

代入a=2005，b=2006，c=2007.
=（1+1+4）/2
3

Answer 3

1.原式= a²+b²-4a+8b+4+16=0
(a²-4a+4)+(b²+8b+16)=0
(a-2)²+(b+4)²=0
(a-2)²≥0 (b+4)²≥0
∴ a-2=0 b+4=0
a=2 b=-4
a+b=-2

2.
原式=1/2(2a²+2b²+2c²-2ab-2bc-2ca)
=1/2(a²-2ab+b²+b²-2bc+c²+a²-2ca+c²)
=1/2[(a-b)²+(b-c)²+(a-c)²]
=1/2(1+1+4)
=3

Answer 4

a²+b²-4a+8b+20=0
(a-2)²+(b+4)²=0
a=2,b=-4
∴a+b=2-4=-2

a²+b²+c²-ab-bc-ca
=1/2（2a²+2b²+2c²-2ab-2bc-2ca）
=1/2[(a-b)²+(b-c)²+(c-a)²]
=1/2(1+1+4)
=3

Answer 5

a²+b²-4a+8b+20=0
则(a-2)^2+(b+4)^2=0
则只有
a-2=0
b+4=0
a=2,b=-4
则a+b=-2

a²+b²+c²-ab-bc-ca
=1/2(2a²+2b²+2c²-2ab-2bc-2ca)
=1/2[(a-b)^2+(b-c)^2+（c-a)^2]
=1/2[1+1+2^2]
=1/2*6
=3