=2ab.[a+2b]+c^2.[a+2b]- c.[a^2+4ab+4.b^2]

=.................................-c[a+2b]^2

=[a+2b].{2ab+c^2-ca-2bc]

=[a+2b]{ 2b.[a-c]-c.[a-c] }

=[a+2b].[a-c].[2b-c]

b) x⁴+2007x²+2006x+2007

=x⁴-x+2007x²+2007x+2007

=x.(x³-1)+2007.(x²+x+1)

=x.(x-1)(x²+x+1)+2007.(x²+x+1)

=(x²+x+1)(x²-x+2007)

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

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

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

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

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

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

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)

phân tích đa thức thành nhân tử 2a^2b + 4ab^2 -a^2c + ac^2 -4b^2c +2bc^2 - 4abc? | Yahoo Hỏi & Đáp

2a²b+4ab²-a²c+ac²-4b²c+2bc²-4abc

=2ab(a+2b)-ac(a+2b)+c²(a+2b)-2bc(a+2b)

=(a+2b)(2ab-ac+c²-2bc)

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

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

b) a³ + b³ + c³ - 3abc

= ( a + b)³ - 3ab - 3ba + c - 3abc

= (a³ + 3a²b + 3ab² + b³) + c³ - (3a²b + 3ab² + 3ab) 

= (a + b)³ + c² - 3ab(a + b + c)

= (a + b + c) [ (a  + b)² - ( a + b )c + c^2 ]  - 3ab(a + b + c)

=  ( a + b + c ) ( a² + b² + 2ab - ac - bc + c² -3ab )

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

Bài làm

a) 2a² + 4ab² - a²c + ac² - 4b²c + 2bc² - 4abc

= ( 2abc+ 4ab² - a²c ) + ( ac² - 4b²c + 2bc² ) - 6abc + 2a²b

= a( 2bc + 4b² - ac ) - c( 2bc + 4b² - ac ) - 6abc + 2a²b

= ( 2bc + 4b² - ac )( a - c ) - ( 2abc - 2a²b ) - 4abc

= ( 2bc + 4b² - ac )( a - c ) - 2ab( c - a ) - 4abc

= ( 2bc + 4b² - ac - 2ab )( a - c ) - 4abc