`b)` `x/3=3/x` Điều kiện: `x\ne0`
`=>x^2=9`
`<=>x^2-9=0`
`<=>(x-3)(x+3)=0`
`<=>`\(\left[ \begin{array}{l}x-3=0\\x+3=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3∈Z(tm)\\x=-3∈Z(tm)\end{array} \right.\)
Vậy `x∈{3;-3}`
`b)` `1/2=x/-2`
`=>2x=-2`
`<=>x=2/-2`
`<=>x=-1(∈Z)`
Vậy `x∈{-1}`