\(a,x^2+6x=x\left(x+6\right)\\ b,9x^2-1=\left(3x\right)^2-1^2=\left(3x-1\right)\left(3x+1\right)\\ c,x^2+2xy-9+y^2=\left(x^2+2xy+y^2\right)-9=\left(x+y\right)^2-3^2=\left(x+y-3\right)\left(x+y+3\right)\\ c,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

a) \(=a\left(a^3-9a^2+a-9\right)=a\left[a^2\left(a-9\right)+\left(a-9\right)\right]\)

\(=a\left(a-9\right)\left(a^2+1\right)\)

b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

c) \(=x\left(2y+z\right)+3\left(2y+z\right)=\left(2y+z\right)\left(x+3\right)\)

d) \(=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-a\right)\left(x-b\right)\)

a) = a(a³-9a²+a-9)

b) =3x²+5y-3xy-5x

= (3x²-5x)+(5y-3xy)

=x(3x-5)+y(5-3x)

=x(3x-5)-y(3x-5)

=(3x-5)(x-y)

c)2xy +3z+6y+xz

=(2xy+6y)+(3z+xz)

=2y(x+3)+z(3+x)

=(x+3)(2y-z)

a) \(a^4-9a^3+a^2-9a=a\left(a^3-9a^2+a-9\right)=a\left[\left(a^3-9a^2\right)+\left(a-9\right)\right]=a\left(a^2+1\right)\left(a-9\right)\)

\(a,x\left(x+6\right)\\ b,\left(9x-1\right)\left(9x+1\right)\\ c,\left(x+y\right)-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ d,\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\)

a) \(4x^2-1\)

\(=\left(2x\right)^2-1^2\)

\(=\left(2x-1\right)\left(2x+1\right)\)

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(9x^2-\dfrac{1}{4}\)

\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)

\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)

d) \(\left(x-y\right)^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

e) \(9-\left(x-y\right)^2\)

\(=3^2-\left(x-y\right)^2\)

\(=\left(3+x-y\right)\left(3-x+y\right)\)

f) \(\left(x^2+4\right)^2-16x^2\)

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

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

\(=\left(x-2\right)^2\left(x+2\right)^2\)

Lời giải:
a.

$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

b. 

$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$

$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$

c.

$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$

$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$

$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$

c.

$x^2-5y^2-y^4+2xy-9$

$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$

\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)

a) x⁸+x⁴+1 = (x⁴+1)²-x⁴ = (x⁴-x²+1)(x⁴+x²+1)

b) x¹²-3x⁶-1 = (x⁶-1)²-x⁶ = (x⁶-x³-1)(x⁶+x³-1)

c) 3x⁴+10x²-25 = 4x⁴-(x⁴-10x²+25) = 4x⁴- (x²-5)² = (x²+5)(3x²-5)

d) x²-5y²-y⁴+2xy-9 = (x+y)²-(y²+3)² = (x+y-y²-3)(x+y+y²+3)

a: \(7x-14y=7\left(x-2y\right)\)

b: \(4x^2-4x+1=\left(2x-1\right)^2\)

\(a,=\left(3x-11\right)\left(3x+11\right)\\ b,=\left(3x+1-x+2\right)\left(3x+1+x-2\right)\\ =\left(2x+3\right)\left(4x-1\right)\\ c,=\left(2x+1-8\right)\left(2x+1+8\right)=\left(2x-7\right)\left(2x+9\right)\)

Câu 2: 

a: =x(x+6)

b: =(3x-1)*(3x+1)

c: \(=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)

d: \(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

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

Bài 1: 

a: \(8x^3-2x=2x\left(4x^2-1\right)=2x\left(2x-1\right)\left(2x+1\right)\)

c: \(-5m^3\left(m+1\right)+m+1=\left(m+1\right)\left(-5m^3+1\right)\)