Question

[monaca]

phân tích đa thức :  \(a^2.\left(b-c\right)+b^2.\left(c-a\right)+c^2.\left(a-b\right)\) thành nhân tử

Answer 1

[Yen Nhi]

`Answer:`

`a^2.(b-c)+b^2.(c-a)+c^2.(a-b)`

`=a^2.(b-c)+b^2[(c-b)-(a-b)]+c^2.(a-b)`

`=a^2.(b-c)+b^2.(c-b)+b^2.(a-b)+c^2.(a-b)`

`=(b-c)(a^2-b^2)-(a-b)(b^2-c^2)`

`=(b-c)(a-b)(a+b)-(a-b)(b-c)(b+c)`

`=(a-b)(b-c)(a+b-b-c)`

`=(a-b)(b-c)(a-c)`

Answer 2

[✎﹏ミ★꧁༺вєѕт↭ℓαυяιєℓ↭νи༻꧂★ミ.༻(Trưở...]

\(a^2.\left(b-c\right)+b^2,\left(c-a\right)+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left(a-c\right)+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left[\left(a-b\right)+\left(b-c\right)\right]+c^2.\left(a-b\right)\)

\(=a^2.\left(b-c\right)-b^2.\left(a-b\right)-b^2.\left(b-c\right)+c^2.\left(a-b\right)\)

\(=\left(b-c\right).\left(a^2-b^2\right)+\left(a-b\right).\left(c^2-b^2\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a+b\right)-\left(b-c\right).\left(b+c\right).\left(a-b\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a+b-b-c\right)\)

\(=\left(b-c\right).\left(a-b\right).\left(a-c\right)\)