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

a²+b²+c²-ab-2c-3b+4=0
a²-ab+b²/4+3b²/4-3b+3+3/4+c²-2c+1=0
(a-b/2)²+3(b/2-1)²+(c-1)²=0
(a-b/2)²=0,3(b/2-1)²=0,(c-1)²=0
a=b/2, b/2=1,c=1
a=b/2, b=2,c=1
a=1
b=2
c=1
c=1
a+b+c
=2+1+1
=4

Answer 3

解：配方即可
a²-ab+b²/4+3b²/4 -3b+3+c²-2c+1=0
(a²-ab+b²/4)+(3/4)(b² -4b+4)+(c²-2c+1)=0
(a-b/2)²+(3/4)(b-2)²+(c-1)²=0
因为平方值都非负
所以 a-b/2=0, b-2=0, c-1=0
所以 b=2, c=1, a=4
所以 a+b+c=7