a) \(A=\dfrac{x}{3+x}+\dfrac{x^2+9}{9-x^2}\left(x\ne\pm3\right).\)
\(A=\dfrac{x}{3+x}+\dfrac{x^2+9}{\left(3-x\right)\left(3+x\right)}=\dfrac{3x-x^2+x^2+9}{\left(3-x\right)\left(3+x\right)}.\)
\(A=\dfrac{3x+9}{\left(3-x\right)\left(3+x\right)}=\dfrac{3\left(x+3\right)}{\left(3-x\right)\left(3+x\right)}=\dfrac{3}{3-x}.\)
\(B=\dfrac{3x+1}{x^2-3x}-\dfrac{1}{x}\left(x\ne0;x\ne3\right).\)
\(B=\dfrac{3x+1}{x\left(x-3\right)}-\dfrac{1}{x}=\dfrac{3x+1-x+3}{x\left(x-3\right)}.\\ B=\dfrac{2x+4}{x\left(x-3\right)}.\)
b) Để \(A\in Z\Leftrightarrow\dfrac{3}{3-x}\in Z.\Rightarrow3-x\inƯ\left(3\right)=\left\{\pm3;\pm1;\right\}\)
\(3-x\) | \(3\) | \(-3\) | \(11\) | \(-1\) |
\(x\) | \(0\) | \(6\) | \(2\) | \(4\) |
TM | TM | TM | TM |
Vậy để \(A\in Z\Leftrightarrow x\in\left\{0;6;2;4\right\}.\)