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26 tháng 6 2019

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

\(=ab^2-ac^2-ba^2+bc^2+ca^2-cb^2\)

\(=\left(ab^2-ac^2-bc^2\right)-\left(ba^2-bc^2-ca^2\right)\)

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

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

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

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

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

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

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

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

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

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

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

phân tích đa thức thành nhân tử

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

= -(b-a)(c-a)(c-b)

nha bạn

30 tháng 8 2021

a2(b-c)+b2(c-a)+c2(a-b)

=a2b-a2c+b2c-b2a+c2(a-b)

=(a2b-b2a)-(a2c-b2c)+c2(a-b)

=ab(a-b)+c(a2-b2)+c2(a-b)

=ab(a-b)+c(a-b)(a+b)+c2(a-b)

=(a-b)(ab+ac+bc+c2)

=(a-b)[(ab+bc)+(ac+c2)]

=(a-b)[b(a+c)+c(a+c)]

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

13 tháng 7 2017

=sure google

7 tháng 8 2018

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

\(=a\left(b-c\right)\left(b+c\right)-bc^2+ba^2+ca^2-cb^2\)

\(=a\left(b-c\right)\left(b+c\right)-\left(bc^2+cb^2\right)+\left(ba^2+ca^2\right)\)

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

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

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

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

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

\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)

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

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

\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)\)

\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)

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

31 tháng 10 2015

\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)

\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)

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

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

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

mình làm vội, có chỗ nào sai bạn thông cảm nha