Question

Tìm giá trị của A:

\(A=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\) biết a + b + c = 4

Answer 1

Áp dụng hằng đẳng thức mở rộng :

\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

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

\(\Leftrightarrow A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2}\)

\(\Leftrightarrow A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2\left(a^2+b^2+c^2-ab-bc-ac\right)}\)

\(\Leftrightarrow A=\dfrac{a+b+c}{2}\)

Thay \(a+b+c=4\) vào M ta có :

\(A=\dfrac{4}{2}=2\)

Vậy A = 2