Question

a²+b²+c²-ab-3b-2c+4=0求a+b+c

Answer 1

a²+b²+c²-ab-3b-2c+4=0
-->
(a²-ab+b²/4)+(3b²/4
-3b)+(c²-2c+1)+3=0
-->
(a-b/2)²+3(b²/4-b+1)+(c-1)²=0
-->
(a-b/2)²+3(b/2-1)²+(c-1)²=0
根据非负数性质得：
a=b/2
,
b/2=1
,
c=1
-->
a=1,b=2,c=1
-->
a+b+c=4
希望能帮助你哈

Answer 2

解：a²+b²+c²-ab-3b-2c+4=0
a²-ab+1/4b²+3/4b²-3b+3+c²-2c+1=0
(a-1/2b)²+3(1/2b-1)²+(c-1)²=0
∵(a-1/2b)²≥0.,3(1/2b-1)²≥0,(c-1)²≥0
∴a-1/2b=0,
1/2b-1=0
c-1=0
解得c=1,b=2.a=1
即a+b+c=4
哪里不清欢迎追问，满意谢谢采纳！