$\text{Đáp án + Giải thích các bước giải:}$
`1//(x-3)(5x+12)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\5x+12=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\5x=-12\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\x=-\dfrac{12}{5}\end{array} \right.\)
`\text{Vậy}` `S={3;-(12)/(5)}`
`2//(2x+1)(2-3x)=0`
`⇔` \(\left[ \begin{array}{l}2x+1=0\\2-3x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=-1\\3x=2\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=\dfrac{2}{3} \end{array} \right.\)
`\text{Vậy}` `S={-(1)/(2);(2)/(3)}`
`3//2x^{2}+5x=0`
`<=>x(2x+5)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\2x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{5}{2}\end{array} \right.\)
`\text{Vậy}` `S={0;-(5)/(2)}`
`4//(x-2019)(x+2020)=0`
`<=>` \(\left[ \begin{array}{l}x-2019=0\\x+2020=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2019\\x=-2020\end{array} \right.\)
`\text{Vậy}` `S={2019;-2020}`
