Question

Cho a,b,c thuộc R thỏa mãn a+b+c=1 Tính GTBT: A=\(\dfrac{a^{3^{ }}+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

Answer 1

Ta có:

** \(a^3+b^3 +c^3 -3abc \)

\(=(a+b)^3+c^3 - 3ab(a+b) - 3abc \)

\(=(a+b+c)[(a+b)^2 - c(a+b)+ c^2] - 3ab(a+b+c) \)

\(=(a+b+c)(a^2 + 2ab+b^2-ca-bc+c^2) - 3ab(a+b+c) \)

\(=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \)

\(=a^2+b^2+c^2-ab-bc-ca\)

** \((a-b)^2 + (b-c)^2+(c- a)^2\)

\(=a^2+b^2+b^2+c^2+c^2+a^2 - 2(ab+bc+ca)\)

\(=2(a^2+b^2+c^2-ab-bc-ca)\)

\(\Rightarrow A=\dfrac{a^2+b^2+c^2-ab-bc-ca}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}=\dfrac{1}{2}\)