Đáp án:

B1

`12a²b-18ab²-30b²`

`=6b(2a²-3ab-5b)`

`27a²(b-1)-9a³(1-b)`

`=27a²(b-1)+9a³(b-1)`

`=(b-1)(27a²+9a³)`

`=(b-1)9a²(3+a)`

`=(b-1)(3+a)(9a²+1)`

B2

`4x(x-2013)-x+2013=0`

`=4x(x-2013)-(x-2013)=0`

`=(x-2013)(4x-1)=0`

`x-2013=0`

`⇒x=2013`

`4x-1=0`

`⇒x=1/4`

`(x+1)²= x+1`

`⇔(x+1)²-(x+1)=0`

`⇔(x+1)(x+1-1)=0`

`x+1=0`

`⇒x=-1`

`x=0`

$\text{*Khiên}$

Giải thích các bước giải:

 

Giải thích các bước giải:

 Bài 1.

\(\begin{split}&12a^2b - 18ab^2 - 30b^2 \\ &=6b\left(2a^2-3ab-5b\right)\end{split}\\ 27a^2(b-1)-9a^3(1-b)\\ \,=27a^2(b-1)+9a^3\left(b-1\right)\\ \,=(b-1)\left(27a^2+9a^3\right)\\ \, =9a^2\left(3+a\right)(b-1)\)

Bài 2.

\(4x(x-2013) - x+2013 =0\\ \Leftrightarrow 4x(x-2013)-(x-2013)=0\\ \Leftrightarrow (x-2013)(4x-1)=0\) \(\Leftrightarrow\left[ \begin{array}{l}x-2013=0\\4x-1=0\end{array} \right.\\ \Leftrightarrow\left[ \begin{array}{l}x=2013\\x=\dfrac 14\end{array} \right.\)

\((x+1)^2= x+1\\\Leftrightarrow(x+1)^2-(x+1)=0\\ \Leftrightarrow (x+1)(x+1-1)=0\\ \Leftrightarrow (x+1)\cdot x=0\\ \Leftrightarrow\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)