Giải thích các bước giải:
a.ĐKXĐ $x+3\ne 0,x^2+x-6\ne 0,2-x\ne 0\to x\ne 2,-3$
b.
Ta có :
$Q=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}$
$\to Q=\dfrac{(x+2)(x-2)}{(x+3)(x-2)}-\dfrac{5}{(x-2)(x+3)}-\dfrac{x+3}{(x-2)(x+3)}$
$\to Q=\dfrac{(x+2)(x-2)-(x-2)(x+3)-x-3}{(x-2)(x+3)}$
$\to Q=\dfrac{-2x-1}{(x-2)(x+3)}$
c.$Q=\dfrac{-3}{2}\to \dfrac{-2x-1}{(x-2)(x+3)}=\dfrac{-3}{2}\to x=\pm\dfrac{-1+\sqrt{241}}{6}$
d.Ta có : $Q\in Z$
$\to \dfrac{-2x-1}{(x-2)(x+3)}\in Z$
$\to \dfrac{(2x+1)^2}{x^2+x-6}\in Z$
$\to \dfrac{4(x^2+x-6)+25}{x^2+x-6}\in Z$
$\to 4+\dfrac{25}{x^2+x-6}\in Z$
$\to\dfrac{25}{x^2+x-6}\in Z$
$\to x^2+x-6\in\{25,5,1,-5,-1,-25\}$
$\to$Không tồn tại x thuộc Z thỏa mãn đề
e.$x^2-9=0\to x= 3(x\ne -3)\to Q=\dfrac{-5}6$
