\(Q=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(Q=\dfrac{a^3+3a^2b+3ab^2+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(Q=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(Q=\dfrac{\left(a+b+c\right).\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(Q=\dfrac{\left(a+b+c\right).\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(Q=\dfrac{\left(a+b+c\right).\left(a^2+b^2-ac-bc+c^2-ab\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(\Rightarrow2Q=\dfrac{\left(a+b+c\right).\left(2a^2+2b^2-2ac-2bc+2c^2-2ab\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(2Q=\dfrac{\left(a+b+c\right).\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)\(2Q=\dfrac{\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(2Q=a+b+c\)
Ta có: \(a+b+c=3\)
\(\Rightarrow2Q=3\)
\(\Rightarrow Q=\dfrac{3}{2}\)
Vậy \(Q=\dfrac{3}{2}\)tại \(a+b+c=3\)
Tham khảo nhé~
