Phân tích đa thức thành nhân tử
a2(b-c)+b2(c-a)+c2(a-b)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(abc-\left(ab+bc+ac\right)+\left(a+b+c\right)-1=\left(abc-ab\right)-\left(bc-b\right)-\left(ac-a\right)+\left(c-1\right)=ab\left(c-1\right)-b\left(c-1\right)-a\left(c-1\right)+\left(c-1\right)=\left(c-1\right)\left(ab-b-a+1\right)=\left(c-1\right)\left[b\left(a-1\right)-\left(a-1\right)\right]=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(B=a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(B=ab^2-ac^2+bc^2-a^2b+a^2c-b^2c\)
\(B=\left(ab^2-a^2b\right)-\left(ac^2-c^2b\right)+\left(a^2c-b^2c\right)\)
\(B=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a-b\right)\left(a+b\right)\)
\(B=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(B=\left(a-b\right)\left[a\left(c-b\right)-c\left(c-b\right)\right]\)
\(B=\left(a-b\right)\left(c-b\right)\left(a-c\right)\)
\(a^3-b^3+c^3+3abc\)
\(=\left(a-b\right)^3+c^3+3abc+3a^2b-3ab^2\)
\(=\left[\left(a-b\right)^3+c^3\right]+3ab\left(c+a-b\right)\)
\(=\left(c+a-b\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(c+a-b\right)\)
\(=\left(c+a-b\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(c+a-b\right)\left(a^2+b^2+c^2+ab-ac+bc\right)\)
\(=\left[a^2+b^2+2-2\left(ab-1\right)\right]\left[a^2+b^2+2+2\left(ab-1\right)\right]\\ =\left(a^2+b^2-2ab+4\right)\left(a^2+b^2+2ab\right)\\ =\left(a+b\right)^2\left(a^2+b^2-2ab+4\right)\)
b) \(a^6-b^3\)
\(=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c) \(x^4-1\)
\(=\left(x^2\right)^2-1^2\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2\left(c-a\right)-c^2\left[\left(b-c\right)+\left(c-a\right)\right]\)
\(=a^2\left(b-c\right)+b^2\left(c-a\right)-c^2\left(b-c\right)-c^2\left(c-a\right)\)
\(=\left(b-c\right)\left(a^2-c^2\right)+\left(c-a\right)\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(a+c\right)+\left(c-a\right)\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(a+c-b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)