Lời giải:

Bài `1` :

`e)`

`6x(x-3y)+2y(3y-x)`

`=6x(x-3y)-2y(x-3y)`

`=(x-3y)(6x-2y)`

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

`f)`

`4xy-2x^(2)+6x`

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

Bài `2` :

`a)`

`5x(x-7)-(x-7)=0`

`⇔(x-7)(5x-1)=0`

$⇔\left[\begin{matrix}x-7=0\\ 5x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=7\\ x=\dfrac{1}{5}\end{matrix}\right.$

Vậy `x=7` hoặc `x=1/5`

`b)`

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

`⇔x(2x+5)+(2x+5)=0`

`⇔(2x+5)(x+1)=0`

$⇔\left[\begin{matrix}2x+5=0\\ x+1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=\dfrac{-5}{2}\\ x=-1\end{matrix}\right.$

Vậy `x=-5/2` hoặc `x=-1`

`c)`

`6x(3x-5)-3x+5=0`

`⇔6x(3x-5)-(3x-5)=0`

`⇔(3x-5)(6x-1)=0`

$⇔\left[\begin{matrix}3x-5=0\\ 6x-1=0\end{matrix}\right.$

$⇔\left[\begin{matrix}x=\dfrac{5}{3}\\ x=\dfrac{1}{6}\end{matrix}\right.$

Vậy `x=5/3` hoặc `x=1/6`

Bài `3` :

`c)`

`3x^(2)+2x=0`

`⇔x(3x+2)=0`

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

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

Vậy `x=0` hoặc `x=-2/3`

Đáp án `+` Giải thích các bước giải `!`

Bài `1:`

`e)`

`6x(x-3y)+2y(3y-x)`

`= 6x(x-3y)-2y(x-3y)`

`= (6x-2y)(x-3y)`

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

`f)`

`4xy-2x^2+6x`

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

Bài `2:`

`a)`

`5x(x-7)-(x-7) = 0`

`<=> (5x-1)(x-7) = 0`

`<=>` \(\left[ \begin{array}{l}5x-1=0\\x-7=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}5x=1\\x=7\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{5}\\x=7\end{array} \right.\) 

Vậy `S= {1/5; 7}`

`b)`

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

`<=> x(2x+5)+(2x+5) = 0`

`<=> (x+1)(2x+5) = 0`

`⇔` \(\left[ \begin{array}{l}x+1=0\\2x+5=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-1\\2x=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-1\\x=-\dfrac{5}{2}\end{array} \right.\) 

Vậy `S= {-1; -(5)/2}`

`c)`

`6x(3x-5)-3x+5 = 0`

`<=> 6x(3x-5)-(3x-5) = 0`

`<=> (6x-1)(3x-5) = 0`

`⇔` \(\left[ \begin{array}{l}6x-1=0\\3x-5=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}6x=1\\3x=5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{6}\\x=\dfrac{5}{3}\end{array} \right.\) 

Vậy `S= {1/6; 5/3}`

Bài `3:`

`c)`

`3x^2+2x = 0`

`<=> x(3x+2) = 0`

`⇔` \(\left[ \begin{array}{l}x=0\\3x+2=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=0\\3x=-2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{2}{3}\end{array} \right.\) 

Vậy `S= {0; -(2)/3}`