a) \(3xy-6xy^2=3xy\left(1-2y\right)\)

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

c) \(x^3-x^2+2\)

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

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

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

g) \(6x^2-12x=6x\left(x-2\right)\)

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

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

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

l) \(x^3-7x^2+9x+3x^2-21x+27=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)=\left(x+3\right)\left(x^2-7x+9\right)\)

sau bạn đăng tách ra cho mn cùng giúp nhé 

a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)

b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)

c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)

d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)

e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)

f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)

a) 5xy ( x - y ) - 2x + 2y

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

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

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

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

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

c)  x² + 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

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

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

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

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

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

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

Bài 10 :

Câu a :

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

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

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

Câu b :

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

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

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

Câu c :

\(x^2+4x-xy-4y\)

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

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

Câu d :

\(3xy+2z-6y-xz\)

\(=\left(3xy-6y\right)-\left(xz-2z\right)\)

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

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

Bài 11 :

Câu a :

\(4-9x^2=0\)

\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)

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

Vậy ........................

Câu b :

\(x^2+x+\dfrac{1}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{2}=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy........................

Câu c :

\(2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy..................

Câu d :

\(3x\left(x-4\right)-x+4=0\)

\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy................................

Câu e :

\(x^3-\dfrac{1}{9}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

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

Vậy........................

Câu f :

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

\(\Leftrightarrow\left(3x-y-x+y\right)\left(3x-y+x-y\right)=0\)

\(\Leftrightarrow2x\left(4x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy..........................

Tìm x:

1. 3x (2x + 3) - (2x + 5).(3x - 2) = 8

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=0 \)

\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)

Vậy x = 5

2. 4x (x -1) - 3(x² - 5) -x² = (x - 3) - (x + 4)

\(\Leftrightarrow4x^2-4x-3x^2+15-x^2=x-3-x-4\)

\(\Leftrightarrow-4x+15=-7\)

\(\Leftrightarrow-4x=-22\Leftrightarrow x=\frac{11}{2}\)

Vậy x = \(\frac{11}{2}\)

3. 2 (3x -1) (2x +5) - 6 (2x - 1) (x + 2) = -6

\(\Leftrightarrow2\left(6x^2+15x-2x-5\right)-6\left(2x^2+4x-x-2\right)=-6\)

\(\Leftrightarrow12x^2+30x-4x-10-12x^2-24x+6x+12=-6\)

\(\Leftrightarrow8x=-8\Leftrightarrow x=-1\)

Vậy x = -1

4. 3 ( 2x - 1) (3x - 1) - (2x - 3) (9x - 1) - 3 = -3

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-18x^2+2x+27x-3-3=-3\)

\(\Leftrightarrow18x^2-6x-9x+3-18x^2+2x+27x-6=-3\)

\(\Leftrightarrow14x=0\Leftrightarrow x=0\)

Vậy x = 0

5. (3x - 1) (2x + 7) - ( x + 1) (6x - 5) = (x + 2) - (x - 5)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=7\)

\(\Leftrightarrow18x=9\Leftrightarrow x=\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)

6. 3xy (x + y) - (x + y) (x² + y² + 2xy) + y³ = 27

\(\Leftrightarrow3x^2y+3xy^2-\left(x+y\right)^3+y^3=27\)

\(\Leftrightarrow3x^2y+3xy^2-x^3-y^3-3x^2y-3xy^2+y^3=27\)

\(\Leftrightarrow-x^3=27\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

7. 3x (8x - 4) - 6x (4x - 3) = 30

\(\Leftrightarrow24x^2-12x-24x^2+12x=30\)

\(\Leftrightarrow0=30\) ( vô lý)

Vậy pt vô nghiệm

8. 3x (5 - 2x) + 2x (3x - 5) = 20

\(\Leftrightarrow15x-6x^2+6x^2-10x=20\)

\(\Leftrightarrow5x=20\Leftrightarrow x=4\)

Vậy x = 4