Đáp án:+Giải thích các bước giải:
a) ĐKXĐ: $\begin{cases} x-1 \ne 0\\x+1\ne 0\\2x\ne0\\5x-5\ne0\\x+2x+1\ne0\end{cases}$ `=>` $\begin{cases}x\ne0\\x \ne \pm1 \end{cases}$
b) `P=((x+1)/(x-1)-(x-1)/(x+1)):(2x)/(5x-5)-(x^2-1)/(x^2+2x+1)`
`=( (x+1)^2/((x+1)(x-1))-(x-1)^2/((x+1)(x-1))):(2x)/(5(x-1))-((x+1)(x-1))/(x+1)^2`
`=((x+1)^2-(x-1)^2)/((x+1)(x-1)) . (5(x-1))/(2x) - (x-1)/(x+1)`
`= (4x)/(x+1). 5/(2x) - (x-1)/(x+1)`
`=10/(x+1) - (x-1)/(x+1)`
`=(10-(x-1))/(x+1) = (11-x)/(x+1)`
c) `P = 2 => (11-x)/(x+1) = 2`
`<=> 11-x = 2(x+1)`
`<=> 11-x = 2x + 2`
`<=> - 3x=-9 <=> x = 3` (TMĐK)
d) `P=(11-x)/(x+1)=(12-(x+1))/(x+1)=12/(x+1)-1`
x nguyên, để `P \in ZZ` thì `12/(x+1) \in ZZ`
`=> x + 1 \in Ư(12)={\pm1;\pm2;\pm3\pm4;\pm6;\pm12}`
`=> x \in {-2;0;1;-3;2;-4;3;-5;5;-7;11;-13}`
mà `x\ne 0; x \ne +-1`
`=> x\ne {+-2;+-3;-4;\pm5;-7;11;-13}`
