a) 6x² - 12x

= 6x(x - 2)

b) x² + 2x + 1 - y²

= (x² + 2x + 1) - y²

= (x + 1)² - y²

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

c) x + y + z + x² + xy + xz

= (x + x²) + (y + xy) + (z + xz)

= x(1 + x) + y(1 + x) + z(1 + x)

= (x + y + z)(x + 1)

d) xy + xz + y² + yz

= (xy + xz) + (y² + yz)

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

= (x + y)(x + z)

e) x³ + x² + x + 1

= (x³ + x²) + (x + 1)

= x²(x + 1) + (x + 1)

= (x² + 1)(x + 1)

f) xy + y - 2x - 2

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

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

= (y - 2)(x + 1)

g) x³ + 3x - 3x² - 9

= (x³ - 3x²) + (3x - 9)

= x²(x - 3) + 3(x - 3)

= (x² + 3)(x - 3)

h) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

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

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

i) 7x² - 7xy - 5x = 5y

mk thấy con này sai sai ý

à câu í là :7x^2-7xy-5x+5y đấy bạn

$\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}$

$=\dfrac{5(x+2)}{4(x-2)}\cdot\dfrac{-2(x-2)}{x+2}$

$=-\dfrac{10}{4}$

$=-\dfrac{5}{2}$

$\dfrac{6x^2y^3}{8x^3y^2}$

$=\dfrac{3y}{4x}$

$\dfrac{x^3-x}{3x+3}$

$=\dfrac{x(x^2-1)}{3(x+1)}$

$=\dfrac{x(x-1)(x+1)}{3(x+1)}$

$=\dfrac{x(x-1)}{3}$

$\dfrac{x^2+3xy}{x^2-9y^2}$

$=\dfrac{x(x+3y)}{(x-3y)(x+3y)}$

$=\dfrac{x}{x-3y}$

$\dfrac{x^2+4x+4}{3x+6}$

$=\dfrac{(x+2)^2}{3(x+2)}$

$=\dfrac{x+2}{3}$

1)(x+1)(x+2)(x+3)=x³+6x²+11x+6

2)a)3x(x-5)-(x-1)(2+3x)=30

<=>3x²-15x-3x²+x+2=30

<=>-14x+2=30

<=>-14x=30-2

<=>-14x=28

<=>x=-2

b)(x+2)(x+3)-(x-2)(x+5)=0

<=>x²+5x+6-x²-3x+10=0

<=>2x+16=0

<=>2x=-16

<=>x=-8

c)(3x+2)(2x+9)-(x+2)(6x+1)=9

<=>6x²+31x+18-6x²-13x-2=9

<=>18x+16=9

<=>18x=9-16

<=>18x=-7

<=>x=-7/18

a, mình nghĩ đề là cm đẳng thức nhé 

\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)

Vậy ta có đpcm 

b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)

\(=-5y-9+xy=VP\)

Vậy ta có đpcm 

c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)

Vậy ta có đpcm 

$P=\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}$

$=\dfrac{2(x-3)+(x+3)}{(x+3)(x-3)}+\dfrac{9-x}{(3-x)(3+x)}$

$=\dfrac{3x-3}{x^2-9}+\dfrac{9-x}{9-x^2}$

$=\dfrac{3x-3}{x^2-9}-\dfrac{9-x}{x^2-9}$

$=\dfrac{3x-3-9+x}{x^2-9}$

$=\dfrac{4x-12}{x^2-9}$

$=\dfrac{4(x-3)}{(x-3)(x+3)}$

$=\dfrac{4}{x+3}$

Vậy $P=\dfrac{4}{x+3}$.

$(2x^4-3x^3-3x^2+6x-1):(x^2-2)$

$=2x^2-3x+1+\dfrac{4x+1}{x^2-2}$

Vậy thương là $2x^2-3x+1$, dư $4x+1$.

$(15x^4y^6-12x^3y^4-18x^2y^3):(-6x^2y^2)$

$=\dfrac{15x^4y^6}{-6x^2y^2}+\dfrac{-12x^3y^4}{-6x^2y^2}+\dfrac{-18x^2y^3}{-6x^2y^2}$

$=-\dfrac52x^2y^4+2xy^2+3y$

Vậy kết quả là $-\dfrac52x^2y^4+2xy^2+3y$.

Bài 3:

a: \(\frac{x}{x-3}+\frac{9-6x}{x^2-3x}\)

\(=\frac{x}{x-3}+\frac{-6x+9}{x\left(x-3\right)}\)

\(=\frac{x^2-6x+9}{x\left(x-3\right)}=\frac{\left(x-3\right)^2}{x\left(x-3\right)}=\frac{x-3}{x}\)

b: \(\frac{6x-3}{x}:\frac{4x^2-1}{3x^2}\)

\(=\frac{3\left(2x-1\right)}{x}\cdot\frac{3x^2}{\left(2x-1\right)\left(2x+1\right)}=\frac{3\cdot3x}{2x+1}=\frac{9x}{2x+1}\)

Bài 2:

a: \(\frac{x^3-x}{3x+3}\)

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

b: \(\frac{x^2+3xy}{x^2-9y^2}=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}=\frac{x}{x-3y}\)

Bài 1:

a: \(\frac{x^2-9}{2x+6}:\frac{3-x}{2}\)

\(=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\frac{2}{-\left(x-3\right)}=\frac{-2}{2}=-1\)

b: \(\frac{2x}{x-y}-\frac{2y}{x-y}=\frac{2x-2y}{x-y}=\frac{2\left(x-y\right)}{x-y}=2\)

c: \(\frac{x+15}{x^2-9}+\frac{2}{x+3}\)

\(=\frac{x+15}{\left(x-3\right)\left(x+3\right)}+\frac{2}{x+3}\)

\(=\frac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+15+2x-6}{\left(x-3\right)\left(x+3\right)}=\frac{3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{3}{x-3}\)

d: \(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{y^2+x^2}{y^2-x^2}\)

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

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

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