Lời giải:

$B=4ab(a-4b)+4b^2c-16bc^2+4ac^2-16a^2c+63abc$

$=4ab(a-4b)-bc(a-4b)+abc+4c^2(a-4b)-4ac^2+4ac^2-16ac(a-4b)-abc$

$=4ab(a-4b)-bc(a-4b)+4c^2(a-4b)-16ac(a-4b)$
$=(a-4b)(4ab-bc+4c^2-16ac)$

$=(a-4b)[b(4a-c)-4c(4a-c)]=(a-4b)(4a-c)(b-4c)$

\(4a^2-4a+1-4b^2\)

<=>\(\left(2a-1\right)^2-4b^2\)

<=>\(\left(2a-1+2b\right)\left(2a-1-2b\right)\)

\(4a^2-4a+1-4b^2\)

\(=\left(2a-1\right)^2-4b^2\)

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

 4a^{2 -} 4b² - 4a -1 
= (4a²- 4a +1 ) - 4b²
= [(2a)² -2a.1 + 1² ] - (2b)2
= (2a -1 )² - (2b)2
= 2a - 1 - 2b ) . ( 2a - 1 + 2b )

a: =x^2+2xy+y^2-4x^2y^2

=(x+y)^2-(2xy)^2

=(x+y+2xy)(x+y-2xy)

b: =49-(a^2-2ab+b^2)

=49-(a-b)^2

=(7-a+b)(7+a-b)

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

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

d: 

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

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

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

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

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

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

(b+a)(c+2a)(c+2b)

=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]