\(a)P=\dfrac{2x}{x+3}+\dfrac{x}{x-3}-\dfrac{2x^2}{x-9}.\left(x\ne\pm3\right).\\ P=\dfrac{2x^2-6x+x^2+3x-2x^2}{\left(x+3\right)\left(x-3\right)}.\\ P=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}.\\ P=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}.\\ P=\dfrac{x}{x+3}\)
\(\Rightarrow\) Điều phài chứng minh.
\(b)\) Ta có: \(P=\dfrac{x}{x+3}=1-\dfrac{3}{x+3}.\)
Để \(P\in Z.\Leftrightarrow1-\dfrac{3}{x+3}\in Z.\Leftrightarrow x+3\in\)\(Ư\) \(\left(3\right)=\left\{1;-1;3;-3\right\}.\)
\(\Rightarrow x\in\left\{-2;-4;0;-6\right\}.\)
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