Đáp án:
$a)A=\dfrac{-6}{x}$
$b)$
$x=-1⇒A=6$
$x=-2⇒A=3$
Giải thích các bước giải:
$\textrm{Ôi dào bạn ơi, Easy!}$
$a)$
$A=\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}$
$=\dfrac{(x-3)^2}{x(x-3)}-\dfrac{x^2}{(x-3)x}+\dfrac{9}{(x-3)x}$
$=\dfrac{(x-3)^2-x^2+9}{x(x-3)}$
$=\dfrac{x^2-6x+9-x^2+9}{x(x-3)}$
$=\dfrac{-6x+18}{x(x-3)}$
$=\dfrac{-(6x-18)}{x(x-3)}$
$=\dfrac{-6(x-3)}{x(x-3)}$
$=\dfrac{-6}{x}$
$b)$
$|x+\dfrac{3}{2}|=\dfrac{1}{2}$
$⇔\left[ \begin{array}{l}x+\dfrac32=-1\\x+\dfrac32=\dfrac{-1}2\end{array} \right.$
$⇔\left[ \begin{array}{l}x=-1\\x=-2\end{array} \right.$
$\textrm{Với}$
$x=-1⇒A=6$
$x=-2⇒A=3$
