$a)ĐKXĐ:x+3\neq0→x\neq-3$
$x-2\neq0→x\neq2$
`b)A=(x+2)/(x+3)-5/((x+3)(x-2))`
`A=((x+2)(x-2))/((x+3)(x-2))-5/((x+3)(x-2))`
`A=(x^2-2^2-5)/((x+3)(x-2))`
`A=(x^2-9)/((x+3)(x-2))`
`A=((x-3)(x+3))/((x+3)(x-2))`
`A=(x-3)/(x-2)`
`c)A=(x-3)/(x-2)`
`→A=(x-2-1)/(x-2)`
`→A=1-1/(x-2)`
`→A∈Z↔1/(x-2)∈Z`
`x∈Z→x-2∈Z→1/(x-2)∈Z→x-2∈Ư(1)={±1}`
`→` \(\left[ \begin{array}{l}x-2=1\\x-2=-1\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
Vậy `x∈{1;3}`
