Đáp án + Giải thích các bước giải:
`a)` `(x-3)(2x+1)=0`
`<=>`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={3;-1/2}`
`b)` `(5x-4)(4x+6)=0`
`<=>`\(\left[ \begin{array}{l}5x-4=0\\4x+6=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{4}{5}\\x=-\dfrac{6}{4}=-\dfrac{3}{2}\end{array} \right.\)
Vậy `S={4/5;-3/2}`
`c)` `(5x-10)(8-2x)=0`
`<=>`\(\left[ \begin{array}{l}5x-10=0\\8-2x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}5x=10\\2x=8\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\)
Vậy `S={2;4}`
`d)` `(9-3x)(15+3x)=0`
`<=>`\(\left[ \begin{array}{l}9-3x=0\\15+3x=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x=9\\3x=-15\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
Vậy `S={3;-5}`.