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

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

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

\(d)4x^2-9y^2+4x+1=\left(2x+1\right)^2-9y^2=\left(2x+3y-1\right)\left(2x-3y+1\right)\)

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

b,x² -y² +4-4x 

=(x²  -4x +4)-y²  

=(x-2)² -y² 

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

Ta có : 6x² - 11x + 3 

= 6x² - 2x - 9x + 3

= (6x² - 2x) - (9x - 3)

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

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

K MIK NHA BẠN !!!!!!!!!!

bÀI 1 

bÀI 2 : 

Bài 3 :

Bài 4: 

5,

6, 

7, 

8,

9, 

10,

11,

12,

13,

K MIK NHA BẠN !!!!!!!!!!

Nguyễn Quang Trung Linh Nhi : Mik cảm ơm 2 bạn nhiều nhóe ;) 

a)   x³ -2x² +5x-4

=x³-x²-x²+x+4x-4

=x²(x-1)-x(x-1)+4(x-1)

=(x²-x+4)(x-1)

b) x³-x²+x+3

=x³+x²-2x²-2x+3x+3

=x²(x+1) -2x(x+1)+3(x+1)

=(x²-2x+3)(x+1)

c) 6x³+x²+x+1

=6x³+ 3x²-2x²-x+2x+1

=6x²(x+\(\frac{1}{2}\)) - 2x(x+\(\frac{1}{2}\)) +2(x+\(\frac{1}{2}\))

=(6x²-2x+2) (x+\(\frac{1}{2}\))

=2( 3x²-x+1) (x+\(\frac{1}{2}\))

d)  4x³ + 6x²+4x+1

= 4x³+2x²+4x²+2x+2x+1

= 4x²(x+\(\frac{1}{2}\))+ 4x(x+\(\frac{1}{2}\))+2(x+\(\frac{1}{2}\))

= 2(2x² +2x+1)( x+\(\frac{1}{2}\))

e) x⁶ -9x³+8

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.

Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết

a) \(x^3-4x^2-12x+27\)

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

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

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

b) \(x^3-3x^2-4x+12\)

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

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

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

a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)

b) \(6x-9-x^2=-\left(x-3\right)^2\)

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

d) \(\left(x-y\right)^2-\left(x+y\right)^2\)

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

\(=2x.\left(-2y\right)=-4xy\)

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

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

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

g) \(x^2+5x-6\)

\(=x^2+6x-x-6\)

\(=x\left(x+6\right)-\left(x+6\right)\)

\(=\left(x-1\right)\left(x-6\right)\)

\(a,=x^2-6x-x+6=\left(x-6\right)\left(x-1\right)\\ b,=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\\ c,=x^3-x^2+2x^2-2x+2x-2=\left(x-1\right)\left(x^2+2x+2\right)\\ d,=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\\ e,=x^2\left(x-6\right)+\left(x-6\right)=\left(x^2+1\right)\left(x-6\right)\\ f,=x^3\left(x^2+1\right)+\left(x^2+1\right)=\left(x^3+1\right)\left(x^2+1\right)\\ =\left(x+1\right)\left(x^2-x+1\right)\left(x^2+1\right)\)

a) \(x^2-7x+6\)

\(=x^2-x-6x+6\)

\(=x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(x-6\right)\left(x-6\right)\)

7(x - 3) - x(3 - x)

= (x - 3)(7 + x)

chỉ bt có v mà k bt có đúng k 

1 ) 7 ( x - 3 ) - x ( 3 - x ) 

= 7 ( x - 3 ) + x ( x - 3 )

= ( x - 3 ) ( 7 + x )

2 ) 4x² - 6x + 3 - 2x

= 4x² - 2x - 6x + 3

= 2x ( 2x - 1 ) - 3 ( 2x - 1 )

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

3 ) ( 4 - x ) - 4x + x²

= ( 4 - x ) - x ( 4 - x )

= ( 4 - x ) (  1 - x )

4 ) x² - 2xy + y²

= ( x - y )²

1, \(7\left(x-3\right)-x\left(3-x\right)=7\left(x-3\right)+x\left(x-3\right)\)

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

2,\(4x^2-6x+\left(3-2x\right)=4x^2-8x+3\)

\(=4x^2-6x-2x+3=2x\left(2x-3\right)-\left(2x-3\right)\)

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

3,\(\left(4-x\right)-4x+x^2=x^2-5x+4\)

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

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

1. P/tích làm sao đc

2. Bạn làm đúng rồi nhưng còn 1 cách:

    từ \(\left(x-2\right)\left(x^2+6x+5\right)=\left(x-2\right)\left(\left(x^2+x\right)+\left(5x+5\right)\right)=\left(x-2\right)\left(x\left(x+1\right)+5\left(x+1\right)\right)=\left(x-2\right)\left(x+1\right)\left(x+5\right)\)