a) 2x^2-3x(x+1)=0 b) 12x(6x-1)-24x^2=0 c) 2x^4+18x^2=0 d (x-2)^2-5(x-2)=0

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`a) 2x^2 -3x(x+1)=0`

`<=> 2x^2 -3x^2 -3x =0`

`<=> -x^2 -3x =0`

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

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

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

Vậy `S={0;-3}`

`----`

`b) 12x(6x-1)-24x^2 =0`

`<=> 72x^2 -12x -24x^2 =0`

`<=> 48x^2 -12x =0`

`<=> 12x(4x-1)=0`

`<=>`\(\left[ \begin{array}{l}12x=0\\4x-1=0\end{array} \right.\) 

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

Vậy `S={0; 1/4}`

`----`

`c) 2x^4 +18x^2=0`

`<=> 2x^2 (x^2 + 9)=0`

`<=>` \(\left[ \begin{array}{l}2x^2=0\\x^2+9=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=0\\x^2=-9(vô lí)\end{array} \right.\)

Vậy `S={0}`

`----`

`d) (x-2)^2 -5(x-2)=0`

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

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

`<=>`\(\left[ \begin{array}{l}x-2=0\\x-7=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=2\\x=7\end{array} \right.\) 

Vậy `S={2;7}`

~~Hok tốt~~