Đáp án + Giải thích các bước giải:
`a//(x+5)(2x-3)=0`
`⇔` \(\left[ \begin{array}{l}x+5=0\\2x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-5\\x=\frac{3}{2}\end{array} \right.\)
Vậy `S={-5;\frac{3}{2}}`
`b//(x^{2}+1)(6x+3)=0`
`⇔6x+3=0` . Vì `x^{2}+1>0`
`⇔6x=-3`
`⇔x=-(1)/(2)`
Vậy `S={-(1)/(2)}`
`c//((3)/(4)-2)((5)/(3)+1)=0`
`⇔` \(\left[ \begin{array}{l}\frac{3}{4}-2=0\\\frac{5}{3}+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\frac{3}{4}-\frac{8}{4}=0\\\frac{5}{3}+\frac{3}{3}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\frac{-5}{4}=0\text{ Vô lí }\\\frac{8}{3}=0\text{ Vô lí }\end{array} \right.\)
Vậy ` ((3)/(4)-2)((5)/(3)+1)\ne0`
`d//2(x+3)(x-4)=0`
`⇔` \(\left[ \begin{array}{l}x+3=0\\x-4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-3\\x=4\end{array} \right.\)
Vậy `S={-3;4}`
