a:
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(P=\left(\dfrac{x^2+3x}{x^3+3x^2+9x+27}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^3-3x^2+9x-27}\right)\)
\(=\left(\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right)\)
\(=\dfrac{x\left(x+3\right)+3\left(x+3\right)}{\left(x^2+9\right)\left(x+3\right)}:\left(\dfrac{1}{x-3}-\dfrac{6x}{\left(x^2+9\right)\left(x-3\right)}\right)\)
\(=\dfrac{x+3}{x^2+9}:\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(=\dfrac{x+3}{x^2+9}\cdot\dfrac{\left(x^2+9\right)\left(x-3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
c: Đặt P=1
=>x+3=x-3
=>3=-3(loại)
Đặt P=0
=>x+3=0
=>x=-3(loại)
Vậy: Khi x>0 thì P không nhận giá trị 1
c: Để P nguyên thì \(x+3⋮x-3\)
=>\(x-3+6⋮x-3\)
=>\(6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{4;2;5;1;6;0;9;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;1;6;0;9\right\}\)
