a: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(A=\left(\dfrac{x}{x+3}-\dfrac{2x}{3-x}+\dfrac{3x^2+9}{9-x^2}\right):\dfrac{3}{x-3}\)
\(=\left(\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{x-3}{3}\)
\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-9}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x-3}{3}\)
\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x+3\right)}\cdot\dfrac{1}{3}\)
\(=\dfrac{3x-9}{3\left(x+3\right)}=\dfrac{x-3}{x+3}\)
b: Để P là số nguyên thì \(x-3⋮x+3\)
=>\(x+3-6⋮x+3\)
=>\(-6⋮x+3\)
=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;-4;-1;-5;0;-6;-9\right\}\)
Lời giải:
a. ĐKXĐ: $x\neq \pm 3$
\(A=\left[\frac{x(x-3)}{(x+3)(x-3)}+\frac{2x(x+3)}{(x-3)(x+3)}-\frac{3x^2+9}{(x-3)(x+3)}\right].\frac{x-3}{3}\\ =\frac{x(x-3)+2x(x+3)-(3x^2+9)}{(x-3)(x+3)}.\frac{x-3}{3}\\ =\frac{3(x-3)}{(x-3)(x+3)}.\frac{x-3}{3}=\frac{x-3}{x+3}\)
b.
$A=\frac{x-3}{x+3}=1-\frac{6}{x+3}$
Để $A$ nguyên thì $\frac{6}{x+3}$ nguyên.
Với $x$ nguyên, điều này xảy ra khi $6\vdots x+3$
$\Rightarrow x+3\in \left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$
$\Rightarrow x\in \left\{-2; -4; -1; -5; 0; -6; 3; -9\right\}$
Do $x\neq \pm 3$ nên $x\in \left\{-2; -4; -1; -5; 0; -6; -9\right\}$
