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26 tháng 3 2020

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

20 tháng 8 2015

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)

11 tháng 8 2015

=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]

20 tháng 3 2016

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 

10 tháng 8 2017

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

20 tháng 8 2015

bạn nên viết ra 2 câu 1 bài

28 tháng 6 2017

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

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

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

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

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

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

AH
Akai Haruma
Giáo viên
20 tháng 10 2020

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

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

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

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

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$

17 tháng 8 2020

Lời giải:

a)

$yz(y+z)+xz(z-x)-xy(x+y)=yz(y+z)+xz^2-x^2z-x^2y-xy^2$

$=yz(y+z)+x(z^2-y^2)-x^2(z+y)$

$=yz(y+z)+x(z-y)(z+y)-x^2(z+y)$

$=(y+z)(yz+xz-xy-x^2)$

$=(y+z)[z(x+y)-x(x+y)]=(y+z)(x+y)(z-x)$

b)

$2a^2b+4ab^2-a^2c+ac^2-4b^2c+2bc^2-4abc$

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

$=2ab(a+2b)-ac(a+2b)+c^2(a+2b)-2bc(a+2b)$

$=(a+2b)(2ab-ac+c^2-2bc)$

$=(a+2b)[2b(a-c)-c(a-c)]$

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

c)

$y(x-2z)^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y[(y-2z)+(x-y)]^2+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x-y)^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=y(y-2z)^2+y(x+y)^2-4xy^2+2y(y-2z)(x-y)+8xyz+x(y-2z)^2-2z(x+y)^2$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-4xy(y-2z)+2y(y-2z)(x-y)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)+2y(y-2z)(x-y-2x)$

$=(y-2z)^2(x+y)+(x+y)^2(y-2z)-2y(y-2z)(x+y)$

$=(x+y)(y-2z)[(y-2z)+(x+y)-2y]=(x+y)(y-2z)(x-2z)$