Lời giải :
BT1 :
`a)`
`3x-6xy`
`=3x(1-2y)`
`b)`
`48x^(3)y^(3)-32x^(2)y^2`
`=16x^(2)y^(2)(3xy-2)`
`c)`
`27x^(3)-8`
`=(3x)^(3)-2^3`
`=(3x-2)(9x^(2)+6x+4)`
`d)`
`125a^(3)-27b^3`
`=(5a)^(3)-(3b)^3`
`=(5a-3b)(25a^(2)+15ab+9b^2)`
`e)`
`3x-3y+x^(2)-2xy+y^2`
`=3(x-y)+(x-y)^2`
`=(x-y)(3+x-y)`
`f)`
`x^(2)-10x+25-4y^2`
`=(x-5)^(2)-(2y)^2`
`=(x-5+2y)(x-5-2y)`
BT2 :
`a)`
`2(x+3)-x(3+x)=0`
`⇔2(x+3)-x(x+3)=0`
`⇔(x+3)(2-x)=0`
$⇔\left[\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=-3\\x=2\end{matrix}\right.$
Vậy `x∈{-3;2}`
`b)`
`x(2x-7)-4x+14=0`
`⇔x(2x-7)-2(2x-7)=0`
`⇔(2x-7)(x-2)=0`
$⇔\left[\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.$
Vậy `x∈{7/2;2}`
`c)`
`x^(3)+4x=0`
`⇔x(x^(2)+4)=0`
Vì `x^2 ≥ 0`
`-> x^(2) + 4 > 0 ∀ x`
`⇔x=0`
Vậy `x=0`
`d)`
`(x+1)^2=x+1`
`⇔x^(2)+2x+1=x+1`
`⇔x^(2)+2x+1-x-1=0`
`⇔x^(2)+x=0`
`⇔x(x+1)=0`
$⇔\left[\begin{matrix}x=0\\x+1=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=0\\x=-1\end{matrix}\right.$
Vậy `x∈{0;-1}`
