ĐK: \[\left\{ {\begin{array}{*{20}{c}}{{x^2} + 3x - 4 \ne 0}\\{x + 4 \ne 0}\\{1 - x \ne 0}\end{array}} \right.\] hay \[\left\{ {\begin{array}{*{20}{c}}{(x - 1)(x + 4) \ne 0}\\{x \ne 1}\\{x \ne - 4}\end{array}} \right.\]. Khi đó \[\left\{ {\begin{array}{*{20}{c}}{x \ne 1}\\{x \ne - 4}\end{array}} \right.\].

\(P = \frac{{10x}}{{{x^2} + 3x - 4}} - \frac{{2x - 3}}{{x + 4}} + \frac{{x + 1}}{{1 - x}}\)

\( = \frac{{10x}}{{(x - 1)(x + 4)}} - \frac{{2x - 3}}{{x + 4}} - \frac{{x + 1}}{{x - 1}}\)

\( = \frac{{10x - (2x - 3)(x - 1) - (x + 1)(x + 4)}}{{(x - 1)(x + 4)}}\)

\( = \frac{{10x - 2{x^2} + 2x + 3x - 3 - {x^2} - 4x - x - 4}}{{(x - 1)(x + 4)}}\)

\( = \frac{{ - 3{x^2} + 10x - 7}}{{(x - 1)(x + 4)}} = \frac{{ - (x - 1)(3x - 7)}}{{(x - 1)(x + 4)}}\)

\( = \frac{{ - 3x + 7}}{{x + 4}}\).

Khi đó \(P = \frac{{ - 3x + 7}}{{x + 4}}\) với \(x \ne 1;\,\,x \ne - \,4\) nên

\(P + 1 = \frac{{ - 3x + 7}}{{x + 4}} + 1 = \frac{{ - 3x + 7 + x + 4}}{{x + 4}} = \frac{{ - 2x + 11}}{{x + 4}} = - 2 + \frac{{19}}{{x + 4}}\).

Với \(x \in \mathbb{Z}\) để \(P + 1 \in \mathbb{Z}\) thì \((x + 4) \in \)Ư\((19) = {\rm{\{ }} \pm 1;\,\, \pm \,9\} \).

\(x + 4\)	\( - 1\)	\(1\)	\( - 19\)	\(19\)
\(x\)	\( - 5\,\,{\rm{(TM)}}\)	\( - 3\,\,{\rm{(TM)}}\)	\( - 25\,\,{\rm{(TM)}}\)	\(15\,\,{\rm{(TM)}}\)
\(P + 1\)	\( - 21\)	\(17\)	\( - 3\)	\( - 1\)

Vậy \(x \in \{ - 25;\,\, - 5;\,\, - 3;\,\,15\} \) thì \[P + 1 \in \mathbb{Z}\].

Phân thức (5x - 7)/(3x^2 + 6x) xác định khi