1,
a, \(A=\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}+\dfrac{x}{x+3}\) (ĐK: \(x\ne\pm3\))
\(=\dfrac{2x\left(x+3\right)+x\left(x-3\right)-3x^2-9}{x^2-9}\)
\(=\dfrac{2x^2+6x+x^2-3x-3x^2-9}{x^2-9}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b, ĐK: \(x\pm3\)
\(A=\dfrac{2}{x-1}\Leftrightarrow\dfrac{3}{x+3}=\dfrac{2}{x-1}\)\(\Leftrightarrow3x-3=2x+6\)\(\Leftrightarrow x=9\left(TM\right)\)
Vậy với \(x=9\) thì A = \(\dfrac{2}{x-1}\)
2,
a, \(A=\dfrac{x}{x-1}+\dfrac{2x^2}{x^2-1}-\dfrac{x}{x+1}\) (ĐK: \(x\pm1\))
\(=\dfrac{x\left(x+1\right)-x\left(x-1\right)+2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2+x+2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x}{x-1}\)
b, ĐK: \(x\pm1\)
\(A=\dfrac{2x}{x-1}=\dfrac{2x-2+2}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{2}{x-1}=2+\dfrac{2}{x-1}\)
Để \(A\in Z\) \(\Leftrightarrow2+\dfrac{2}{x-1}\in Z\Leftrightarrow\dfrac{2}{x-1}\in Z\Leftrightarrow x-1\inƯ_{\left(2\right)}\)
\(\Leftrightarrow x-1\in\left\{\pm1\right\}\) \(\Leftrightarrow x\in\left\{0;2\right\}\)
Vậy với \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) thì A \(\in Z\)
