`a)`
`(4x+2)(5-x)=0`
`<=>` \(\left[ \begin{array}{l}4x+2=0\\5-x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=5\end{array} \right.\)
Vậy `S={-1/2;5}`
``
`b)`
`1/(x-2)-2/(2-x)=5/((x-1)(x-2))(x\ne1;x\ne2)`
`<=>1/(x-2)+2/(x-2)=5/((x-1)(x-2))`
`<=>(x-1+2(x-1))/((x-1)(x-2))=5/((x-1)(x-2))`
`=>x-1+2(x-1)=5`
`<=>x-1+2x-2=5`
`<=>3x-3=5`
`<=>3x=8`
`<=>x=8/3` (thoả mãn)
Vậy `S={8/3}`