f: Ta có: \(3a^3-15a+15b-3b^3\)

\(=3\left(a^3-5a+5b-b^3\right)\)

\(=3\left[\left(a-b\right)\left(a^2+ab+b^2\right)-5\left(a-b\right)\right]\)

\(=3\left(a-b\right)\left(a^2+ab+b^2-5\right)\)

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

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

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

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

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

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

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

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: \(2x^2-7xy+3y^2+x-3y\)

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

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

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

\(a^2-9-8ab+16b^2\)

\(=a^2-8ab+16b^2-9\)

\(=\left(a-4b\right)^2-9\)

\(=\left(a-4b-3\right)\left(a-4b+3\right)\)

a² - 9 - 8ab + 16b²

⇔ (a² - 8ab + 16b²) - 9

⇔ (a - 4b)² - 3²

⇔ (a - 4b - 3)(a - 4b + 3)

\(=(a^{2}-8ab+16b^{2})-9\)

\(=(a-4b)^{2}-3^{2}\)

\(=(a-4b-3)(a-4b+3)\)

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

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

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

\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)

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

b. \(=\left(1-x\right)^3\)

Bài 2:

1)  \(x^2-4x+4=\left(x-2\right)^2\)

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

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

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

5) \(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}x-8y\right)\left(\dfrac{1}{5}x+8y\right)\)

6) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

Bài 2:

7) \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

8) \(x^3+64=\left(x+4\right)\left(x^2+4x+16\right)\)

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

10) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)=2a\cdot2b=4ab\)

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

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

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

12) \(\left(6x-1\right)^2-\left(3x+2\right)^2=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)=\left(9x+1\right)\left(3x-3\right)\)

1:

1: ,4x^2-6x=2x(2x-3)

2: 9x^3y^2+3x^2y^2=3x^2y^2(3x+1)

3: x^3+2x^2+3x=x(x^2+2x+3)

4: 2x^2-4x=2x(x-2)

5: 3x-6y=3(x-2y)

6: x^2-3x=x(x-3)

7: 6x^2y+4xy^2+2xy

=2xy(3x+2y+1)

8: 5x^2(x-2y)-15x(x-2y)

=(x-2y)(5x^2-15x)

=5x(x-3)(x-2y)

9: =3(x-y)+5y(x-y)

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

10: =(x-1)(3x+5)

11: =2(2x-1)-3(2x-1)

=-(2x-1)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)