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

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

c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)

d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)

a) Ta có: \(x^2-2x-4y^2-4y\)

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

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

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

b) Ta có: \(x^3+2x^2+2x+1\)

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

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

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

d) Ta có: \(a^6-a^4+2a^3+2a^2\)

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

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

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

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

c) Ta có: \(x^3-4x^2+12x-27\)

\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)

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

2a3 – 54b3

= 2(a3 – 27b3)

= 2[a3 – (3b)3]

= 2(a – 3b)(a2 + 3ab + 9b2)

\(2a^2+8b^2-8ab\)

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

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

cám ơn nhaaaaa!!!!

2a2+8b2−8ab

=2(�2−4��+4�2)=2(a2−4ab+4b2)

=2(�−2�)2=2(a−2b)2

\(8a^4-2a^2-4a+2\)

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

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

\(8a^4-2a^2-4a+2\)

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

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

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

Lời giải:

a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$

$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$

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

c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$

d.

$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$

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

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

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

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

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

\(---\)

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

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

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

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

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

\(---\)

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

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

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

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

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

\(a.a^4+a^2+1\) 

\(=\left(a^4+2a^2+1\right)-a^2\) 

\(=\left(a^2+1\right)^2-a^2\)  

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

\(b.a^4+a^2-2\) 

\(=a^4+2a^2-a^2-2\) 

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

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

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

\(c.x^3-5x^2-14x\) 

\(=x^3+2x^2-7x^2-14\) 

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

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

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

b: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

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

\(=\left(x-3y\right)\left(2x-y+1\right)\)

Lời giải:
a.

Đặt $2a^2+5ab-3b^2-7b-2=(a+mb+n)(2a+pb+k)$ với $m,n,p,k$ nguyên 

$\Leftrightarrow 2a^2+5ab-3b^2-7b-2=2a^2+ab(2m+p)+mpb^2+a(k+2n)+b(km+np)+kn$ 

Đồng nhất hệ số:

\(\left\{\begin{matrix} 2m+p=5\\ mp=-3\\ k+2n=0\\ km+np=-7\\ kn=-2\end{matrix}\right.\)

Giải hpt này ta thu được $m=3; n=1; p=-1; k=-2$

Vậy $2a^2+5ab-3b^2-7b-2=(a+3b+1)(2a-b-2)$

b. Đa thức không phân tích được thành nhân tử

lm theo pp đồng nhất hệ số ạ

b: Ta có: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

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

\(=\left(x-3y\right)\left(2x-y+1\right)\)

\(a^6+a^4+a^2b^2+b^4-b^6\\ =a^6-b^6+a^4+a^2b^2+b^4\\ =\left(a^6-b^6\right)+\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2\right)^3-\left(b^2\right)^3\right]+\left(a^4+a^2b^2+b^4\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^2+a^2b^2+b^4\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left(a^2-b^2+1\right)\left(a^4+2a^2b^2+b^4-a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left[\left(a^2+b^2\right)^2-\left(ab\right)^2\right]\\ =\left(a^2-b^2+1\right)\left(a^2+b^2-ab\right)\left(a^2+b^2+ab\right)\)

a6, a4 là số mũ hay hệ số vậy bn