a, ĐKXĐ của biểu thức là :
x\(\ne2\) và x\(\ne-3\)
b, Rút gọn A :
\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{x^2-2x+3x-6}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x^2-2x\right)+\left(3x-6\right)}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{5}{x\left(x-2\right)+3\left(x-2\right)}-\dfrac{1}{x-2}=\dfrac{x+2}{x+3}-\dfrac{2}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x+2\right)\left(x-2\right)-5-x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{1}{x+3}\)
c, Ta có
A=\(\dfrac{1}{x+3}=-\dfrac{3}{4}\Leftrightarrow\dfrac{1.4}{\left(x+3\right)4}=\dfrac{-3\left(x+3\right)}{\left(x+3\right).4}\Leftrightarrow4=-3\left(x+3\right)\Leftrightarrow4=-3x-9\Leftrightarrow3x=-9-4\Leftrightarrow3x=-13\Rightarrow x=-\dfrac{13}{3}\left(TM\text{Đ}K\text{X}\text{Đ}\right)\)
