Question

数学 已知a^2+b^2+c^2-ab-3b-2c+4=0,求的a+b+c值

Answer 1

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

所以a+b+c=1+2+1=4

Answer 2

即(a^2-ab+b^2/4)+3(b^2/4-b+1)+(c^2-2c+1)=0
(a-b/2)^2+3(b/2-1)^2+(c-1)^2=0
平方相加为0则都等于0
所以a-b/2=0,b/2-1=0,c-1=0
所以b=2
a=b/2=1
c=1
所以a+b+c=4

Answer 3

a^2+b^2+c^2-ab-3b-2c+4=0
a^2-ab+b^2/4+3b^2/4-3b+3+c^2-2c+1=0
(a-b/2)^2+3(1-b/2)^2+(c-1)^2=0
1-b/2=0 b=2
a-b/2=0 a=1
c-1=0 c=1
a+b+c=1+2+1=4

Answer 4

(0a-b/2)^2+3(b/2-1)^2+(c-1)^2=0
a-b/2=0
b/2-1=0
c-1=0
即a=1
b=2
c=1
∴a+b+c=4