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`a, 2(x-3)=x^2-9`

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

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

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

`<=> [(x-3=0),(x+1=0):}`

`<=> [(x=3),(x=-1):}`

Vậy `S={3;-1}`

`b, (2-3x)^2-4x^2=0`

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

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

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

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

\(⇔\left[ \begin{array}{l}x=\dfrac25\\x=2\end{array} \right.\)

Vậy `S={2;2/5}`

`c, (x^2-4)^2 - (x-2)(x+2)=0`

`<=> (x^2-4)^2 - (x^2-4) = 0`

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

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

`<=> [(x-2=0),(x+2=0),(x^2-5=0):}`

`<=> [(x=2),(x=-2),(x^2=5):}`

`<=> [(x=2),(x=-2),(x=+-\sqrt5):}`

Vậy `S={+-2;+-\sqrt5}`

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

`a)`

`2(x-3) = x^2-9`

`<=> 2x-6-x^2+9 = 0`

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

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

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

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

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

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

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

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

Vậy `S= {-1; 3}`

`b)`

`(2-3x)^2-4x^2 = 0`

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

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

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

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

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

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

Vậy `S= {2/5; 2}`

`c)`

`(x^2-4)^2-(x-2)(x+2) = 0`

`<=> x^4-8x^2+16-x^2+4 = 0`

`<=> x^4-9x^2+20 = 0`

`<=> x^4-5x^2-4x^2+20 = 0`

`<=> (x^4-5x^2)-(4x^2-20) = 0`

`<=> x^2(x^2-5)-4(x^2-5) = 0`

`<=> (x^2-4)(x^2-5) = 0`

`⇔` \(\left[ \begin{array}{l}x^2-4=0\\x^2-5=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x^2=4\\x^2=5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=2\\x=-2\\x=\sqrt{5}\\x=-\sqrt{5}\end{array} \right.\) 

Vậy `S= {+-2; +-\sqrt{5}}`