$a,$ $2x(x-3)-3(x-3)=0$

$⇔(x-3)(2x-3)=0$

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

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

$⇔$ \(\left[ \begin{array}{l}x=\dfrac{3}{2} \\x=3\end{array} \right.\) 

Vậy `S={\frac{3}{2};3}`

$b,$ $x^2(x-1)+4(1-x)=0$

$⇔x^2(x-1)-4(x-1)=0$

$⇔(x-1)(x^2-4)=0$

$⇔$ \(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\) 

Vậy `S={1;±2}`

$c,$ $2x(x-5)=(x-5)^2$

$⇔2x(x-5)-(x-5)^2=0$

$⇔(x-5)(2x-x+5)=0$

$⇔(x-5)(x+5)=0$

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

Vậy `S={±5}`

$d,$ $(2x-1)^2=(4-3x)^2$

$⇔$ \(\left[ \begin{array}{l}2x-1=4-3x\\2x-1=3x-4\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}5x=5\\-x=-3\end{array} \right.\) 

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

Vậy `S={1;3}`

$e,$ $2x(3-4x)-5x^2(4x-3)=0$

$⇔2x(3-4x)+5x^2(3-4x)=0$

$⇔(2x+5x^2)(3-4x)=0$

$⇔x(5x+2)(3-4x)=0$

$⇔$ \(\left[ \begin{array}{l}x=0\\5x+2=0 \\ 3-4x=0 \end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=0\\x=-\dfrac{2}{5} \\ x=\dfrac{3}{4} \end{array} \right.\) 

Vậy `S={0;-\frac{2}{5};\frac{3}{4}}`

Giải thích các bước giải:
 a ↔ 2x²-6-3x+9=0
    ↔ 2x²-3x+3=0
    theo pt ta có a=2 , b=-3, c=3
    Δ = (-3)²-4 . 2.3 = -15
  ⇒ pt vô nghiệm