Đáp án:

Câu `1` :

`a)`

`x^(2)(5x^(3)-x-6)`

`=x^(2).5x^(3)-x^(2).x-x^(2).6`

`=5x^(5)-x^(3)-6x^2`

`b)`

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

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

`=(x-y)^3`

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

Câu `2` :

`a)`

`y^(2)+2y+1`

`=y^(2)+2.y.1+1^2`

`=(y+1)^2`

`b)`

`9x^(2)+y^(2)-6xy`

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

`=(3x)^(2)-2.3x.y+y^2`

`=(3x-y)^2`

`c)`

`25a^(2)+4b^(2)+20ab`

`=25a^(2)+20ab+4b^2`

`=(5a)^(2)+2.5a.2b+(2b)^2`

`=(5a+2b)^2`

`d)`

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

`=x^(2)-2.x.(1)/2+(1/2)^2`

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

Câu `3` :

`a)`

`14x^(2)y-21xy^(2)+28x^(2)y^2`

`=7xy.2x-7xy.3y+7xy.4xy`

`=7xy.(2x-3y+2xy)`

`b)`

`27x^(3)-1/27`

`=(3x)^(3)-(1/3)^3`

`=(3x-1/3)[(3x)^(2)+3x.(1)/3+(1/3)^2]`

`=(3x-1/3)(9x^(2)+x+1/9)`

`c)`

`3x^(2)-3xy-5x+5y`

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

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

`d)`

`x^(2)+7x+12`

`=x^(2)+3x+4x+12`

`=x(x+3)+4(x+3)`

`=(x+3)(x+4)`

Câu `4` :

`a)`

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

`⇔x(x-2)+(x-2)=0`

`⇔(x-2)(x+1)=0`

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

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

Vậy `x=2` hoặc `x=-1`

`b)`

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

`⇔5x(x-3)-(x-3)=0`

`⇔(x-3)(5x-1)=0`

$⇔\left[\begin{matrix}x-3=0\\ 5x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=3\\ 5x=1\end{matrix}\right.$

$⇔\left[\begin{matrix}x=3\\ x=\dfrac{1}{5}\end{matrix}\right.$

Vậy `x=3` hoặc `x=1/5`