a: =(16x+20)^2-(10x+10)^2

=(16x+20-10x-10)(16x+20+10x+10)

=(26x+30)(6x+10)

=4(13x+15)(3x+5)

b: =(x-y+4-2x-3y+1)(x-y+4+2x+3y-1)

=(-x-4y+5)(3x+2y+3)

c: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

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

=2(x^2+1)*4x

=8x(x^2+1)

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

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

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

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

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

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

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

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

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

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

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

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

1:

a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)

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

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

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)

a)\(\frac{1}{64}x^6-125y^3=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

b)\(x^6+1=\left(x^2\right)^3+1^3=\left(x^2+1\right)\left(x^4+x^2+1\right)\)

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

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

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

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

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

1.A

2.C

3.B

4.C

a

c

b

c

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

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

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

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

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

c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)

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

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