\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)

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

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

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

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

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

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

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

a) \(=\left(2x-1\right)^2\)

b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

a. \(4x^2-4x+1=\left(2x\right)^2-2x.2.1+1^2=\left(2x-1\right)^2\)

b. \(xy^2-x^3+2x^2-x=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x^2-2x+1\right)\right]=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

a) \(4\left(x+1\right)^3-x-1=4\left(x+1\right)^3-\left(x+1\right)=\left(x+1\right)\left[4\left(x+1\right)^2-1\right]=\left(x+1\right)\left[2\left(x+1\right)-1\right]\left[2\left(x+1\right)+1\right]=\left(x+1\right)\left(2x+1\right)\left(2x+3\right)\)

b) \(5x\left(x-3\right)+\left(3-x\right)^2-\left(x-3\right)=5x\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)=\left(x-3\right)\left(5x+x-3-1\right)=\left(x-3\right)\left(6x-4\right)=2\left(x-3\right)\left(3x-2\right)\)

c) \(9x^2y^3-3x^4y^2-6x^3y^2+16xy^4=xy^2\left(9xy-3x^3-6x^2+16y^2\right)\)

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)

b) \(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)=25\Leftrightarrow1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)-25=0\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)\left(1+xy\right)+\left(1+xy\right)^2-25=0\Leftrightarrow\left(x+y+1+xy\right)^2-25=0\Leftrightarrow\left(x+y+xy-24\right)\left(x+y+xy+26\right)=0\)

a: Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)

\(=\left(x^2+3x+1\right)^2\)

A = abc - (ab + bc + ca) + a + b + c - 1

= (abc - ab) - (bc - b) - (ac - a) + (c - 1)

= ab(c - 1) - b(c - 1) - a(c - 1) + (c - 1) 

= (ab - b - a + 1)(c - 1) 

= (a - 1).(b - 1).(c - 1)   

a: \(=\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+4x+1\right)\)