\(a,3x\left(x+1\right)+9\left(x+1\right)\\ =\left(3x+9\right)\left(x+1\right)\\ =3\left(x+3\right)\left(x+1\right)\\ b,x^2-5xy+2x-10y\\ =x\left(x-5y\right)+2\left(x-5y\right)\\ =\left(x+2\right)\left(x-5y\right)\)

BÀI 2: Phân tích thành nhân tử .

`a, 3x(x+1)+9(x+1)`

`= (x+1)(3x+9)`

`=3(x+1)(x+3)`

`b,x^2-5xy+2x-10y`

`=x(x-5y)+2(x-5y)`

`=(x-5y)(x+2)`

a. 6x³-x²-486x+81

= 6x³-54x²+53x²-477x-9x+81

= 6x².(x-9)+53x.(x-9)-9.(x-9)

= (x-9).(6x²+53x-9)

= (x-9)(6x²+54x-x-9)

=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)

b. x³-5x²+3x+9

= x³+x²-6x²-6x+9x+9

=x².(x+1)-6x.(x+1)+9.(x+1)

=(x+1)(x²-6x+9)=(x+1)(x-3)²

c. x³+3x²+6x+4

= x³+x²+2x²+2x+4x+4

= x².(x+1)+2x.(x+1)+4.(x+1)

= (x+1)(x²+2x+4)

d. 

(x - 4)(x² + 4x + 16) - x(x² - 6) = 2

x³ - 64 - x³ + 6x = 2

6x = 2 + 64

6x = 66

x = 66 : 6

x = 11

x³ - 27 + 3x(x - 3)

= (x - 3)(x² + 3x + 9) + 3x(x - 3)

= (x - 3)(x² + 3x + 9 + 3x)

= (x - 3)(x² + 6x + 9)

= (x - 3)(x + 3)²

5x³ - 7x² + 10x - 14

= 5x(x² + 2) - 7(x² + 2)

= (x² + 2)(5x - 7)

a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

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

c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

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

a ) ( 3x² + 3x + 2)² - ( 3x² + 3x - 2)²

=(3x² + 3x + 2 + 3x² + 3x - 2) [( 3x² + 3x + 2) - ( 3x² + 3x - 2) ]

=(6x²+6x)*4

=24x(x+1)

b ) ( xy+1)² - ( x+y)²

=( xy+1 + x+y ) [( xy+1) - ( x+y)]

=[x(y+1)+(y+1)] [x(y-1) - (y-1)] 

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

c ) ( x + y)³ - ( x - y)³

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

=2y( x²+2xy+y² - x²+y²+ x²-2xy +y² )

=2y(3y²+x²)

d ) 4( x² - y² ) - 8(x - ay) - 4(a² - 1)

=4(-a²+2ay-y²+x²-2x+1)

=4[-(a-y)²+(x-1)²]

=-4(y-x-a+1)(y+x-a-1)

bài 1:

a. x² - 5=0

=>x² = 0+5 = 5

=> x = \(\sqrt{5}\)

vậy x= \(\sqrt{5}\)

sorry biết mỗi a thôi

a) x² - 5 = 0

    x²       = 0 + 5

    x²        = 5

=> x = \(\sqrt{5}\)

Vậy ...

\(x^2-5=0\)

<=>  \(\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

<=>  \(\orbr{\begin{cases}x-\sqrt{5}=0\\x+\sqrt{5}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\)

a) x.(1-x)+(x-1)²        

⁼x.1-x²+x²-2.x².1+1²

⁼x-x²+x²-2.x²+1

=(-x²+x²-2x²)+1

=-2x²+1

b)(x+1)²-3.(x+1)

=x²+2.x².1+1²-3.x+3

=x²+2.x²+1-3x+3

=(x²+2x²)+(1+3)-3x

=3x²-3x+4

c)3x.(x-1)²-(1-x)³

=3x.x²-2,x².1+1²-1³-3.1².x+3.x.1²=3x.x²-2x²+1-1-3x+3x=(3x-3x+3x)(x²-2x²)(1-1)=3x.(-x²)