`a)2x(x-3)-3(x-3)=0`
`⇔(2x-3)(x-3)=0`
`⇔` \(\left[ \begin{array}{l}2x-3=0\\x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=3\end{array} \right.\)
Vậy `S={3/2;3}`
`b)x^2(x-1)+4(1-x)=0`
`⇔(x^2-4)(x-1)=0`
`⇔(x-2)(x+2)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x+2=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-2\\x=1\end{array} \right.\)
Vậy `S={2;-2;1}`
`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`
`⇔` \(\left[ \begin{array}{l}x-5=0\\x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\)
Vậy `S={5;-5}`
`d)(2x-1)^2=(4-3x)^2`
`⇔(2x-1)^2-(4-3x)^2=0`
`⇔(2x-1-4+3x)(2x-1+4-3x)=0`
`⇔(5x-5)(-x+3)=0`
`⇔` \(\left[ \begin{array}{l}5x-5=0\\-x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
Vậy `S={1;3}`
`e)2x(3-4x)-5x^2(4x-3)=0`
`⇔(-2x-5x^2)(4x-3)=0`
`⇔x(-2-5x)(4x-3)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\-2-5x=0\\4x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{2}{5}\\x=\dfrac{3}{4}\end{array} \right.\)
Vậy `S={0;-2/5;3/4}`
