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22 tháng 2 2020

\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2+8abc\)

\(=a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)+8abc\)

\(=ab^2-2abc+ac^2+bc^2-2abc+ba^2+ca^2-2abc+cb^2+8abc\)

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

\(=\left(ac^2+bc^2\right)+\left(ab^2+ba^2\right)+\left(ca^2+cb^2+2abc\right)\)

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

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

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

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

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

11 tháng 3 2021

Ta có: 

\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)

\(A=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)

\(A=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)

\(A=\left(2c+1\right)\left(4ab+2a+2b+1\right)\)

\(A=\left(2c+1\right)\left[2a\left(2b+1\right)+\left(2b+1\right)\right]\)

\(A=\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)\)

11 tháng 3 2021

Ta có:\(A=8abc+4\left(ab+bc+ca\right)+2\left(a+b+c\right)+1\)

\(=8abc+4ab+4bc+4ca+2a+2b+2c+1\)

\(=\left(8abc+4ab\right)+\left(4bc+2b\right)+\left(4ca+2a\right)+\left(2c+1\right)\)

\(=4ab\left(2c+1\right)+2b\left(2c+1\right)+2a\left(2c+1\right)+\left(2c+1\right)\)

\(=\left(2c+1\right)\left(4ab+2b+2a+1\right)\)

\(=\left(2c+1\right)\left[2b\left(2a+1\right)+\left(2a+1\right)\right]\)

\(=\left(2c+1\right)\left(2b+1\right)\left(2a+1\right)\)

23 tháng 8 2023

�=8���+4(��+��+��)+2(�+�+�)+1

A = 8abc + 4ab + 4bc + 4ca + 2a + 2b + 2c + 1

�=(8���+4��)+(4��+2�)+(4��+2�)+(2�+1)

�=4��(2�+1)+2�(2�+1)+2�(2�+1)+(2�+1)

�=(2�+1)(4��+2�+2�+1)

�=(2�+1)[2�(2�+1)+(2�+1)]

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)

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

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

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

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

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

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

12 tháng 5 2020

tk mình đi mình giải cho 

26 tháng 6 2019

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

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

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

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

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

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

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

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

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

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

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

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

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

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

20 tháng 11 2018

Mình không biết

20 tháng 11 2018

ko bt thì  ko nói nha mình đang cần gấp lém xin đừng trêu