\(9a^3-13a+6=\left(9a^3-6a^2\right)+\left(6a^2-4a\right)-\left(9a-6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)

\(x^4-4x^3+8x+3=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(5x^2+5x\right)+\left(3x+3\right)=x^3\left(x+1\right)-5x^2\left(x+1\right)+5x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^3-5x^2+5x+3\right)=\left(x+1\right)\left[\left(x^3-3x^2\right)-\left(2x^2-6x\right)-\left(x-3\right)\right]=\left(x+1\right)\left(x-3\right)\left(x^2-2x-1\right)\)

\(a^4-9a^3+81a-81\)

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

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

\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)

a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)

b: \(ax^2+a-a^2x-x\)

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

\(=\left(x-a\right)\left(ax-1\right)\)

\(a,a^4+a^2+1\)

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

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

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

\(---\)

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

\(=a^4-a^2+2a^2-2\)

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

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

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

\(---\)

\(c,x^3-5x^2-14x\)

\(=x^3+2x^2-7x^2-14x\)

\(=x^2\left(x+2\right)-7x\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-7x\right)\)

\(=x\left(x+2\right)\left(x-7\right)\)

\(a.a^4+a^2+1\) 

\(=\left(a^4+2a^2+1\right)-a^2\) 

\(=\left(a^2+1\right)^2-a^2\)  

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

\(b.a^4+a^2-2\) 

\(=a^4+2a^2-a^2-2\) 

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

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

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

\(c.x^3-5x^2-14x\) 

\(=x^3+2x^2-7x^2-14\) 

\(=x^3\left(x+2\right)-7x\left(x+2\right)\) 

\(=\left(x^3-7x\right)\left(x+2\right)\) 

\(=x\left(x-7x\right)\left(x+2\right)\) 

\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

\(=c\left(a-b\right)^2+\left[ab^2+ac^2+a^2b+bc^2-a^3-b^3-c^3\right]\)

\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)+ab^2+a^2b-a^3-b^3\)

\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)-\left(a^3-a^2b\right)+\left(ab^2-b^3\right)\)

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

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

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

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

a: =x^3(x-y)+(x-y)

=(x-y)(x^3+1)

=(x-y)(x+1)(x^2-x+1)

b: =(a-1)^2-9b^2

=(a-1-3b)(a-1+3b)

=a⁴-81-(9a³-81a)

=(a²-9)(a²+9) -9a(a²-9)

=(a²-9)(a²-9a+9)

ĐẾN ĐÂY MÌNH CHỊU ,BẠN GIẢI TIẾP NHA

a)\(3xy-6y=3y\left(x-2\right)\)

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

c)\(y^2-81=y^2-9^2=\left(y-9\right)\left(y+9\right)\)

a)`3xy-6y`

`=3y(x-2)`

b)`x(x+y)+2x+2y`

`=x(x+y)+2(x+y)`

`=(x+y)(x+2)`

c)`y^2 -81`

`=y^2-9^2`

`=(y-9)(y+9)`

a) \(25a^2-1=\left(5a-1\right)\left(5a+1\right)\)

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

c) \(\dfrac{1}{4}a^2-\dfrac{9}{25}=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)\)

d) \(\dfrac{9}{4}a^4-\dfrac{16}{25}=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\)

e) \(\left(2a+b\right)^2-a^2=\left(2a+b-a\right)\left(2a+b+a\right)=\left(a+b\right)\left(3a+b\right)\)

f) \(16\left(x-1\right)^2-25\left(x+y\right)^2=\left(4x-4-5x-5y\right)\left(4x-4+5x+5y\right)=\left(-x-4-5y\right)\left(9x+5y-4\right)\)

a/ $25x^2-1\\=(5x)^2-1^2\\=(5x-1)(5x+1)$

b/ $a^2-9\\=a^2-3^2\\=(a-3)(a+3)$

c/ $\dfrac{1}{4}a^2-\dfrac{9}{25}\\=\left(\dfrac{1}{2}a\right)^2-\left(\dfrac{3}{5}\right)^2\\=\left(\dfrac{1}{2}a-\dfrac{3}{5}\right)\left(\dfrac{1}{2}a+\dfrac{3}{5}\right)$

d/ $\dfrac{9}{4}a^4-\dfrac{16}{25}\\=\left(\dfrac{3}{2}a^2\right)^2-\left(\dfrac{4}{5}\right)^2\\=\left(\dfrac{3}{2}a^2-\dfrac{4}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left[\left(\sqrt{\dfrac 3 2}a\right)^2-\left(\dfrac{2\sqrt 5}{5}\right)^2\right]\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)\\=\left(\sqrt{\dfrac 3 2}a-\dfrac{2\sqrt 5}{5}\right)\left(\sqrt{\dfrac 3 2}a+\dfrac{2\sqrt 5}{5}\right)\left(\dfrac{3}{2}a^2+\dfrac{4}{5}\right)$

e/ $(2a+b)^2-a^2\\=(2a+b-a)(2a+b+a)\\=(a+b)(3a+b)$

f/ $16(x-1)^2-25(x+y)^2\\=[4(x-1)]^2-[5(x-y)]^2\\=[4(x-1)-5(x-y)][4(x-1)+5(x-y)]\\=[4x-4-5x+5y][4x-4+5x-5y]\\=(-x+5y-4)(9x-5y-4)$

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) \(=mp\left(m^2+mn-mp-np\right)=mp\left[m\left(m+n\right)-p\left(m+n\right)\right]=mp\left(m+n\right)\left(m-p\right)\)

b) \(=abm^2+abn^2+a^2mn+b^2mn=am\left(bm+an\right)+bn\left(bm+an\right)\)

\(=\left(bm+an\right)\left(am+bn\right)\)