Đáp án:
`a) ĐKXĐ:`
$\left\{\begin{matrix}x+3 \ne 0& \\x-2 \ne 0& \end{matrix}\right.$
`->` $\left\{\begin{matrix}x \ne -3& \\x \ne 2& \end{matrix}\right.$
`b)`
`A=(x+2)/(x+3)-5/((x+3)(x-2))`
`MTC: (x+3)(x-2)`
`A=((x+2)(x-2))/((x+3)(x-2))-5/((x+3)(x-2))`
`A=(x^2-4)/((x+3)(x-2))-5/((x+3)(x-3))`
`A=(x^2-4-5)/((x+3)(x-2))`
`A=(x^-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)=(x-2-1)/(x-2)=1-1/(x-2)`
Để `A` nguyên thì `1/(x-2)` nguyên
`-> x-2 \in Ư(1)={-1, 1}`
`-> x \in{1; 3}`
