a.
\(P=\dfrac{x^2-6x+9}{9-x^2}+\dfrac{4x+8}{x+3}=\dfrac{\left(x-3\right)^2}{-\left(x-3\right)\left(x+3\right)}+\dfrac{4x+8}{x+3}\)
\(=\dfrac{x-3}{-\left(x+3\right)}+\dfrac{4x+8}{x+3}=\dfrac{3-x}{x+3}+\dfrac{4x+8}{x+3}\)
\(=\dfrac{3x+11}{x+3}\)
b.
\(x=7\Rightarrow P=\dfrac{3.7+11}{7+3}=\dfrac{32}{10}\)
c.
\(P=\dfrac{3x+11}{x+3}=\dfrac{3\left(x+3\right)+2}{x+3}=3+\dfrac{2}{x+3}\)
\(P\in Z\Rightarrow\dfrac{2}{x+3}\in Z\Rightarrow x+3=Ư\left(2\right)\)
\(\Rightarrow x+3=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x=\left\{-5;-4;-2;-1\right\}\)
`a)`
`ĐKXĐ:x ne +-3`
`P = (x^2 - 6x + 9)/(9-x^2) + (4x+8)/(x+3)`
`=-((x-3)^2)/(x^2 - 9) + (4x+8)/(x+3)`
`=(-(x-3)^2)/((x-3)(x+3)) + (4x+8)/(x+3)`
`=(-x+3)/(x+3) + (4x+8)/(x+3)`
`=(-x+3+4x+8)/(x+3)`
`=(3x+11)/(x+3)`
`b)`
Thay `x=7` vào `P` được `:`
\(P = \dfrac{{3.7 + 11}}{{7 + 3}} = \dfrac{{16}}{5}\)
`c)`
Để `P in Z`
`=>3x+11 vdots x+3`
`=>3x+11 vdots 3x+9`
`=>3x+11 - 3x - 9 vdots x+3`
`=>2 vdots x+3`
`=>x+3 in Ư(2)`
`=>x+3 in {1;2;-1;-2}`
`=>x in {-2;-1;-4;-5}`
Vậy `x in {-2;-1;-4;-5}`
