a/ ĐKXĐ: x khác -1
\(P=\left(\dfrac{4}{x+1}-1\right):\dfrac{9-x^2}{x^2+2x+1}=\left(\dfrac{4}{x+1}-\dfrac{x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}\)
\(=\dfrac{3-x}{x+1}\cdot\dfrac{\left(x+1\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{x+1}{x+3}\)
b/ |x + 1| = 2
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
Với x = 1 P = \(\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
c/ \(\dfrac{x+1}{x+3}=\dfrac{x+3-2}{x+3}=\dfrac{x+3}{x+3}-\dfrac{2}{x+3}=1-\dfrac{2}{x+3}\)
ĐỂ P nguyên thì \(\dfrac{2}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(2\right)\)
\(x+3=\left\{-2;-1;1;2\right\}\)
=> \(x=\left\{-5;-4;-2;-1\right\}\) (tm)
Vậy............
