Lời giải:

a.

$ab(a-b)+bc(b-c)+ca(c-a)$

$=ab(a-b)-bc[(a-b)+(c-a)]+ca(c-a)$

$=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)$

$=(a-b)(ab-bc)-(c-a)(bc-ca)=b(a-b)(a-c)-c(c-a)(b-a)$

$=b(a-b)(a-c)-c(a-c)(a-b)=(a-b)(b-c)(a-c)$

b.

$x^2-3xy-10y^2=(x^2+2xy)-(5xy+10y^2)$

$=x(x+2y)-5y(x+2y)=(x+2y)(x-5y)$

c.

$3x(x-2)-x+2=3x(x-2)-(x-2)=(x-2)(3x-1)$

\(a,ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ =a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)\\ =\left(a^2b-bc^2\right)-\left(ab^2-b^2c\right)+ca\left(c-a\right)\\ =b\left(a-c\right)\left(a+c\right)-b^2\left(a-c\right)-ca\left(a-c\right)\\ =\left(a-c\right)\left(ab+bc-b^2-ca\right)\\ =\left(a-c\right)\left(b-c\right)\left(a-b\right)\)

\(b,x^2-3xy-10y^2\\ =x^2+2xy-5xy-10y^2\\ =x\left(x+2y\right)-5y\left(x+2y\right)=\left(x-5y\right)\left(x+2y\right)\)

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

b: Ta có: \(x^2-3xy-10y^2\)

\(=x^2-5xy+2xy-10y^2\)

\(=x\left(x-5y\right)+2y\left(x-5y\right)\)

\(=\left(x-5y\right)\left(x+2y\right)\)

c: Ta có: \(3x\left(x-2\right)-x+2\)

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

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

a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

a: =(x-2)(3x-2)

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

b: \(=x\left(x^2+2xy+y^2-4\right)\)

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

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

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

Lời giải:
a.

$x^2-7x+6=(x^2-x)-(6x-6)=x(x-1)-6(x-1)=(x-1)(x-6)$

b.

$x-3\sqrt{3}x-12\sqrt{3}$ không phân tích được thành nhân tử

c.

$x^2+4x-2$ không phân tích được thành nhân tử với các hệ số nguyên.

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

\(=xy\left(x^2-2xy+5\right)\)

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

\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

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

\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

A = x² + 3x + 2

A = x² + x + 2x + 2

A = x(x + 1) + 2(x + 1)

A = (x + 2)(x + 1)

a.

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

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)

b.

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

\(=\left(x-y\right)\left(x-y-z\right)\)

c.

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

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

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

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

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

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

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

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

\(=\left(x-y\right)\left(x-y-z\right)\)

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

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

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

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