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Bạn tham khảo bài.
`a+b+c=0`
`-> (a+b)^3 +c^3=0`
`-> a^3 + b^3 + c^3 + 3ab (a+b)=0`
`->a^3 +b^3+c^3 - 3abc=0`
`-> (a+b)^3+c^3 -3ab (a+b)-3abc=0`
`-> (a+b+c)[(a+b)^2 - (a+b)c +c^2] - 3ab(a+b+c)=0`
`-> (a+b+c)(a^2 +b^2 +c^2 - ab-bc-ac)=0`
Do đó : `a^2 + b^2+c^2 - ab-bc-ac=0`
`-> 2a^2 +2b^2+c^2-2ab-2bc-2ac=0`
`-> (a^2 - 2ab+b^2)+(b^2 - 2bc+c^2)+(c^2 -2ca +a^2)=0`
`->(a-b)^2 + (b-c)^2 + (c-a)^2=0`
Vì `(a-b)^2 + (b-c)^2 + (c-a)^2 ≥0∀a,b,c`
Dấu "`=`" xảy ra khi :
`(a-b)^2=0,(b-c)^2=0,(c-a)^2=0`
`↔ a-b=0,b-c=0,c-a=0`
`↔a=b=c`
`A=a^2/(bc)+b^2/(ca)+c^2/(ab)`
`->A=a^2/a^2 + b^2/b^2 +c^2/c^2`
`->A=1+1+1`
`->A=3`
Vậy `A=3`
