Question

Phân tích thành nhân tử: 

\(2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc\)

Answer 1

2a^2b + 4ab^2 -a^2c + ac^2 -4b^2c +2bc^2 - 4abc 
= (2a^2b - 4abc + 2bc^2) + (4ab^2 - 4b^2c) - (a^2c - ac^2) 
= 2b(a^2 - 2ac + c^2) + 4b^2(a - c) - ac(a - c) 
= 2b(a - c)^2 + 4b^2(a - c) - ac(a - c) 
= (a - c)  [ 2b(a - c) + 4b^2 - ac ] 
= (a - c) (2ab -2bc +4b^2 - ac) 
= (a - c)  [ (2ab - ac) + (4b^2 - 2bc) ] 
= (a - c) [a(2b - c) + 2b(2b - c)] 
= (a - c)(2b - c)(a + 2b)

Answer 2

TL:

=\(\left(2a^2b-4bc+2bc^2\right)+\left(4ab^2-4b^2c\right)-\left(a^2c-ac2\right)\) 

=\(2b\left(a^2-2c+c^2\right)+4b^2\left(a-c\right)-ac\left(a-c\right)\) 

=\(2b\left(a-c\right)+4b^2\left(a-c\right)-ac\left(a-c\right)\) 

=\(\left(a-c\right)\left(2b+4b^2-ac\right)\)

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