D.3(x-2y)

đáp án D 

k cho mik nha

Kết quả phân tích đa thức 3x - 6y thành nhân tử là:

A. 3(x - 6y)

B. 3(3x - y)

C. 3(3x - 2y)

D. 3(x - 2y)

x³ - 3x²y + 3xy² - y³ - z³

= (x³ - 3x²y + 3xy² - y³) - z³

= (x - y)³ - z³

= (x - y - z)[(x - y)² + (x - y)z + z²]

= (x - y - z)(x² - 2xy + y² + xz - yz + z³)

--------------------

x² - y² + 8x + 6y + 7

= (x² + 8x + 16) - (y² - 6y + 9)

= (x + 4)² - (y - 3)²

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

= (x - y + 7)(x + y + 1)

a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)

\(=\left(x-y\right)^3-z^3\)

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

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

b: \(=x^2+8x+16-y^2+6y-9\)

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

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

=(x+y+1)(x-y+7)

a) 2x²-2y²-6x-6y = ( 2x²-2y²)- ( 6x+6y)

= 2(x2-y2)- 6(x+y)

2( x+y )- 6( x+y )

..........

b) x^3 +3x^2 - 3x -1= (x^3-1) +(3x^2-3x)

= (x³-1)+ 3x( x-y)

.......

Những dòng mình .... là đến đấy đơn giản rồi, đặt nhân tử chung là đc

= 

#Vũ Khánh Linh câu a/ dòg thứ 3 hình như sai sai.... @@

à ừ mình nhầm để mk sửa

a/ 2x^2-2y^2-6x-6y= ( 2x^2-2y^2)-(6x+6y)

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

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

Chon D

D

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

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

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

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

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

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

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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

a,3x-6y

=3(x-2y)

b,=x^2(2/5+5x+y)

k đúng cho mk nhé bạn

b) 2xy(1+z)

d) 33x^2

f) (x-2y)(3x+y)

b, 2xy+2xyz
=2xy(1+z)

d,27x^2+6x^2
=3x²(9+2)
=3x².11

f, (3x-6y)x+y(x-2y)
=3(x-2y)+y(x-2y)
=(3+y)(x-2y)

b: \(2xy+2xyz=2xy\left(z+1\right)\)

d: \(27x^2+6x^2=33x^2\)

f: \(\left(3x-6y\right)x+y\left(x-2y\right)=\left(x-2y\right)\left(3x+y\right)\)

a, \(\left(3x-6y\right)x+y\left(x-2y\right)=\left(3x+y\right)\left(x-2y\right)\)

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

c, \(\dfrac{2}{5}x\left(y-1\right)+\dfrac{2}{5}y\left(1-y\right)=\left(\dfrac{2}{5}x-\dfrac{2}{5}y\right)\left(y-1\right)=\dfrac{2}{5}\left(x-y\right)\left(y-1\right)\)

d, \(x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\)

a² + b² + 2ab + 2a + 2b + 1

= ( a² + 2ab + b² ) + ( 2a + 2b ) + 1

= ( a + b )² + 2( a + b ) + 1²

= ( a + b + 1 )²

3x( x - 2y ) - 6y( 2y - x )

= 3x( x - 2y ) + 6y( x - 2y )

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

x² + 2x - 3

= x² - x + 3x - 3

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

= ( x - 1 )( x + 3 )

a) \(a^2+b^2+2ab+2a+2b+1\)

\(=\left(a^2+2ab+b^2\right)+\left(2a+2b\right)+1\)

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

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

b) \(3x\left(x-2y\right)-6y\left(2y-x\right)\)

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

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

c) \(x^2+2x-3=x^2-x+3x-3\)

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

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

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