Question

Tìm a,b,c biết

a²-2a+b²+4b+4c²-4c+6=0

Answer 1

 a^2-2a+b^2+4b+4c^2-4c+6=0 
<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0 
<=>(a-1)^2+(b+2)^2+(2c-1)^2=0 
vi (a-1)^2>=0,(b+2)^2>=0,(2c-1)^2>=0 
=>(a-1)^2+(b+2)^2+(2c-1)^2>=0 
dau = xay ra <=>(a-1)^2=0,(b+2)^2=0,(2c-1)^2=0 
<=>a-1=b+2=2c-1=0 
<=>a=2,b=-2,c=1/2 
vay a=2,b=-2,c=1/2

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Answer 2

 a^2-2a+b^2+4b+4c^2-4c+6=0 
<=>(a^2-2a+1)+(b^2+4b+4)+(4c^2-4c+1)=0 
<=>(a-1)^2+(b+2)^2+(2c-1)^2=0 
vi (a-1)^2>=0,(b+2)^2>=0,(2c-1)^2>=0 
=>(a-1)^2+(b+2)^2+(2c-1)^2>=0 
dau = xay ra <=>(a-1)^2=0,(b+2)^2=0,(2c-1)^2=0 
<=>a-1=b+2=2c-1=0 
<=>a=2,b=-2,c=1/2 
vay a=2,b=-2,c=1/2

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