x(2x+4)(3x-1)(2x+4) 2x(5x-2)-3(2-5x)=0

$2x(5x-2)-3(2-5x)=0$

⇔ $2x(5x-2) +3 (5x-2)=0$

⇔ $(2x+3) (5x-2) =0$

⇔ $2x+3 =0 hoặc 5x-2=0$

⇔ $x = 3/2  và x = 2/5$

$x(2x+4)(3x-1)(2x+4)$

⇔ $2x + 4 = 0$

 ⇔$  x        = 0$
 ⇔  $  3x - 1 = 0$
⇔ $x = -2$
⇔       $x = 0$
⇔ $   x =1/3$

Đáp án + Giải thích các bước giải:

 `a )`

` TH_1 : x ( 2x + 4 ) - ( 3x - 1 ) ( 2x + 4 ) = 0 `

`<=> ( 2x + 4 ) ( x - 3x + 1 ) = 0 `

`<=> ( 2x + 4 ) ( - 2x + 1 ) = 0 `

`<=> `\(\left[ \begin{array}{l}2x+4=0\\-2x+1=0\end{array} \right.\) 

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

Vậy ` S = { - 2 ; 1/2 } `

` TH_2 : x ( 2x + 4 ) + ( 3x - 1 ) ( 2x + 4 ) = 0 `

`<=> ( 2x + 4 ) ( x + 3x - 1 ) = 0 `

`<=> ( 2x + 4 ) ( 4x - 1 ) = 0 `

`<=> `\(\left[ \begin{array}{l}2x+4=0\\4x-1=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{4} \end{array} \right.\) 

Vậy `S = { - 2 ; 1/4 } `

`TH_3 : x ( 2x + 4 ) ( 3x - 1 ) ( 2x + 4 ) = 0 `

`<=> x ( 2x + 4 ) ( 3x - 1 ) = 0 `

`<=>`\(\left[ \begin{array}{l}x=0\\2x+4=0\\3x-1=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=0\\2x=-4\\3x=1\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=0\\x=-2\\x=\dfrac{1}{3} \end{array} \right.\)

Vậy `S = { 0 ; - 2 ; 1/2 } ` 

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`b) 2x ( 5x - 2 ) - 3 ( 2 - 5x ) = 0 `

`<=> 2x ( 5x - 2 ) + 3 ( 5x - 2 ) = 0`

`<=> ( 5x - 2 ) ( 2x + 3 ) = 0`

`<=>`\(\left[ \begin{array}{l}5x-2=0\\2x+3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}5x=2\\2x=-3\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=\dfrac{2}{5} \\x=-\dfrac{3}{2} \end{array} \right.\) 

Vậy `S = { 2/6 ; - 3/2 } `