Đáp án:

Bài `1` :

`a)`

`x^(3)+x^(2)-4x-4`

`=x^(2)(x+1)-4(x+1)`

`=(x+1)(x^(2)-4)`

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

`b)`

`x^(2)-y^(2)-4x+4`

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

`=(x^(2)-2.x.2+2^2)-y^2`

`=(x-2)^(2)-y^2`

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

`c)`

`(x^(2)+9)-36x^2`

`=(x^(2)+9)-(6x)^2`

`=(x^(2)-6x+9)(x^(2)+6x+9)`

`=(x^(2)-2.x.3+3^2)(x^(2)+2.x.3+3^2)`

`=(x-3)^(2)(x+3)^2`

`d)`

`25-x^(2)+2xy-y^2`

`=25-(x^(2)-2xy+y^2)`

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

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

`e)`

`x^(3)-4x^(2)+4x-1`

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

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

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

`f)`

`3x-3y-x^(2)+2xy-y^2`

`=3(x-y)-(x^(2)-2xy+y^2)`

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

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

`g)`

`2x^(2)-9x+10`

`=2x^(2)-4x-5x+10`

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

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

`h)`

`x^(2)-5x-14`

`=x^(2)-7x+2x-14`

`=x(x-7)+2(x-7)`

`=(x-7)(x+2)`

`i)`

`x^(3)-3x^(2)+2`

`=x^(3)-x^(2)-2x^(2)+2x-2x+2`

`=x^(2)(x-1)-2x(x-1)-2(x-1)`

`=(x-1)(x^(2)-2x-2)`

`k)`

`x^(4)+4`

`=x^(4)+4x^(2)+4-4x^2`

`=[(x^2)^(2)+2.x^(2).2+2^2]-4x^2`

`=(x^(2)+2)^(2)-(2x)^2`

`=(x^(2)+2-2x)(x^(2)+2+2x)`

Bài `2` :

`a)`

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

`⇔(x-3)(2x+5)=0`

$⇔\left[\begin{matrix}x-3=0\\ 2x+5=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=3\\ 2x=-5\end{matrix}\right.$

$⇔\left[\begin{matrix}x=3\\ x=\dfrac{-5}{2}\end{matrix}\right.$

Vậy `x∈{2;-5/2}`

`b)`

`(x^(2)-4)-(x-2)(3-2x)=0`

`⇔(x-2)(x+2)-(x-2)(3-2x)=0`

`⇔(x-2)(x+2-3+2x)=0`

`⇔(x-2)(3x-1)=0`

$⇔\left[\begin{matrix}x-2=0\\ 3x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=2\\ 3x=1\end{matrix}\right.$

$⇔\left[\begin{matrix}x=2\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `x∈{2;1/3}`

`c)`

`(2x+5)^(2)=(x+2)^2`

`⇔(2x+5)^(2)-(x+2)^2=0`

`⇔(2x+5-x-2)(2x+5+x+2)=0`

`⇔(x+3)(3x+7)=0`

$⇔\left[\begin{matrix}x+3=0\\ 3x+7=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=-3\\ 3x=-7\end{matrix}\right.$

$⇔\left[\begin{matrix}x=-3\\ x=\dfrac{-7}{3}\end{matrix}\right.$

Vậy `x∈{-3;-7/3}`

`d)`

`x^(2)-5x+6=0`

`⇔x^(2)-2x-3x+6=0`

`⇔x(x-2)-3(x-2)=0`

`⇔(x-2)(x-3)=0`

$⇔\left[\begin{matrix}x-2=0\\ x-3=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=2\\ x=3\end{matrix}\right.$

Vậy `x∈{2;3}`

`e)`

`2x^(3)+6x^(2)=x^(2)+3x`

`⇔2x^(3)+6x^(2)-(x^(2)+3x)=0`

`⇔2x^(2)(x+3)-x(x+3)=0`

`⇔(x+3)(2x^(2)-x)=0`

`⇔x(x+3)(2x-1)=0`

$⇔\left[\begin{matrix}x=0\\ x+3\\ 2x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=0\\ x=-3\\ 2x=1\end{matrix}\right.$

$⇔\left[\begin{matrix}x=0\\ x=-3\\ x=\dfrac{1}{2}\end{matrix}\right.$

Vậy `x∈{0;-3;1/2}`

`f)`

`(x^(2)-1)(x+2)-(x-2)(x^(2)+2x+4)-2x^(2)=0`

`⇔x^(3)+2x^(2)-x-2-x^(3)+8-2x^2=0`

`⇔(x^(3)-x^2)+(2x^(2)-2x^2)-x+(-2+8)=0`

`⇔-x+6=0`

`⇔-x=-6`

`⇔x=6`

Vậy `x=6`

Đáp án:

 mk sẽ viết ở giải thích nhé!

Giải thích các bước giải:

 a,x^3+x^2-4x-4=(x^3+x^2)-(4x+4)

=x^2(x+1)-4(x+1)

=(x+1)(x^2-4)

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

b,x^2-y^2-4x+4

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

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

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

c,(x^2+9)^2-36x^2

=(x+9-6x)(x+9+6x)

=(-5x+9)(7x+9)

d,25-x^2+2xy-y^2

=25-(x^2-2xy+y^2)

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

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

e,x^3-4x^2+4x-1

=x^3-(4x^2-4x+1)

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

(bn lm như phép d nhé,áp dg HDT số 3 )

f,3x-3y-x^2+2xy-y^2

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

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

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

g,h mk chx nghĩ ra (thông cảm)

bài 2

a,2x(x-3)+5(x-3)=0

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

suy ra 2 trường hợp

x-3=0 suy ra x=3

2x+5=0 suy ra x=-5/2