$\text{Bài 5:}$
$a) \text{ĐKXĐ:}$
$•x+3 \ne 0 \to x\ne -3$
$• x^2+x-6\ne0$
$\to x^2+3x-2x-6\ne0$
$\to x(x+3)-2(x+6)\ne 0$
$\to (x+3)(x-2)\ne 0$
$\to x\ne 2; -3$
$•2-x\ne 0 \to x\ne 2$
$\to x\ne {-3;2}$
$b) A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}$
$\to A=\dfrac{x+2}{x+3}-\dfrac{5}{(x-2)(x+3)}-\dfrac{1}{x-2}$
$\to A=\dfrac{(x+2)(x-2)-5-(x+3)}{(x-2)(x+3)}$
$\to A=\dfrac{x^2-4-5-x-3}{(x-2)(x+3)}$
$\to A=\dfrac{x^2-x-12}{(x-2)(x+3)}$
$\to A=\dfrac{x^2-4x+3x-12}{(x-2)(x+3)}$
$\to A=\dfrac{x(x-4)+3(x-4)}{(x-2)(x+3)}$
$\to A=\dfrac{(x+3)(x-4)}{(x-2)(x+3)}$
$\to A=\dfrac{x-4}{x-2}$
