Cho biểu thức A=(3x/x-2-2x^2-5/x^2-4-x-1/x+2):3/x+2

Rút gọn  và tìm điều kiện xác định của $A$.

$A=\left(\frac{3x}{x-2}-\frac{2x^2-5}{x^2-4}-\frac{x-1}{x+2}\right):\frac{3}{x+2}$

DK:$\left\{\begin{array}{l}x-2\ne 0\\ x+2\ne 0\iff x\ne \pm 2\\ x^2-4\ne 0\end{array}\right.$

$A=\left(\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2x^2-5}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{3}{x+2}$

$A=\left(\frac{3x^2+6x-2x^2+5-x^2+x+2x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{3}{x+2}$

$A=\frac{9x+3}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3}$

$A=\frac{3\left(3x+1\right)}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3}$

$A=\frac{3x+1}{x-2}$