1/

`a, 6x^2 - 3x`

`= 3x . 2x - 3x . 1`

`= 3x(2x - 1)`

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

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

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

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

2/

`a, 5x^2 - 5 = 0`

`=> 5(x^2 - 1) = 0`

`=> x^2 - 1 = 0`

`=> (x - 1)(x + 1) = 0`

`=> x - 1 = 0` hoặc `x + 1 = 0`

`=> x = 1` hoặc `x = -1`

Vậy `x = 1` hoặc `x = -1`

`b, x^2 - 9x + 14 = 0`

`=> x^2 - 2x - 7x + 14 = 0`

`=> x(x - 2) - 7(x - 2) = 0`

`=> (x - 2)(x - 7) = 0`

`=> x - 2 = 0` hoặc `x - 7 = 0`

`=> x= 2` hoặc `x = 7`

Vậy `x = 2` hoặc `x = 7`

Bài `1:`

$a,\\6x^2-3x\\=3x.2x-2x.1\\=3x(2x-1)\\b,\\x^2-y^2+4x+4\\=(x^2+4x4)-y^2\\=(x^2+2.2.x+2^2)-y^2\\=(x+2)^2-y^2\\=(x+2-y)(x+2+y)$

Bài `2:`

$a,\\5x^2-5=0\\⇔5(x^2-1)=0\\⇔(x-1)(x+1)=0\\⇔\left[\begin{matrix} x-1=0\\ x+1=0\end{matrix}\right.\\⇔\left[\begin{matrix} x=1\\ x=-1\end{matrix}\right.\\\text{Vậy}, x=1;x=-1\\b,\\x^2-9x+14=0\\⇔x^2-2x-7x+14=0\\⇔x(x-2)-7(x-2)=0\\⇔(x-2)(x-7)=0\\⇔\left[\begin{matrix} x-2=0\\ x-7=0\end{matrix}\right.\\⇔\left[\begin{matrix} x=2\\x=7\end{matrix}\right.\\\text{Vậy},x=2;x=7$