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`

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

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

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

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

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

x² - x - y² - y

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

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

***

9x² + y² - 16z² + 6xy

= (3x + y)² - (4z)²

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

***

a³ - a²x - ay + xy

= a²(a - x) - y(a - x)

= (a - x)(a² - y)

***

2x² - 8y² + 3x + 6y

= 2(x² - 4y²) + 3(x + 2y)

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

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

***

xy(x + y) + yz(y + z) + xz(x + z) + 2xyz

= xy(x + y + z) + yz(x + y + z) + xz(x + z)

= y(x + y + z)(x + z) + xz(x + z)

= (x + z)(xy + y² + yz + xz)

= (x + z)[y(x + y) + z(x + y)]

= (x + z)(x + y)(y + 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)  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

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.

a. $x^2-2x+1=(x-1)^2$

b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$

c. $5x^2-10xy=5x(x-2y)$

d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$

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

a) Ta có: \(x^2-y^2-2x+2y\)

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

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

b) Ta có: \(2x+2y-x^2-xy\)

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

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

c) Ta có: \(x^2-25+y^2+2xy\)

\(=\left(x+y\right)^2-25\)

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

d) Ta có: \(3x^2-6xy+3y^2-12z^2\)

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

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

e) Ta có: \(x^2+2xy+y^2-xz-yz\)

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

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

f) Ta có: \(x^2-2x-4y^2-4y\)

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

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

a: =(a^2-b^2)-(2a-2b)

=(a-b)(a+b)-2(a-b)

=(a-b)(a+b-2)

b: =(3x-3y)+5y(x-y)

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

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

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

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

=(x-y)[(x+y)^2-x]

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

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

e: =16-(x^2-4xy+4y^2)

=16-(x-2y)^2

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

g: =9x^2-6x+1-(3xy-y)

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

=(3x-1)(3x-y-1)

h: =(x-y)^3-z^3

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

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

a) \(a^2-b^2-2a+2b\)

\(=\left(a^2-b^2\right)-\left(2a-2b\right)\)

\(=\left(a+b\right)\left(a-b\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2\right)\)

b) \(3x-3y-5x\left(y-x\right)\)

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

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

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

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

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

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

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

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

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

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

a) \(xy+y-2x-2\)

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

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

b) \(xy+1+x+y\)

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

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

c) \(x\left(x-1\right)+y\left(x-1\right)+z\left(x-1\right)\)

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