trả lời gấp nhen mấy bạn mk đang ở ranh giới của sự sống còn]
b ghi đề cho rõ nhé.
\(\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{a^3+b^3+c^3-3abc+3a^2b+3ab^2-3a^2b-3ab^2}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{\left(a+b\right)^3+c^3+3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]+3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-3ab-ac-bc\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{1}{2}.\frac{\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{1}{2}.\frac{\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{1}{2}.\frac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(=\frac{1}{2}.\left(a+b+c\right)\) ( đk: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ne0\))
Tham khảo nhé~
