B có nghĩa khi\(\left\{{}\begin{matrix}2x+3\ge0\\x-3>0\end{matrix}\right.=>\left\{{}\begin{matrix}x\ge\dfrac{-3}{2}\\x>3\end{matrix}\right.=>x>3\)

A có nghĩa khi \(\dfrac{2x+3}{x-3}\ge0=>\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+3\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+3\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{3}{2}\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{3}{2}\\x\le3\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x\ge3\\x\le\dfrac{-3}{2}\end{matrix}\right.\)

\(x^2\) = y =>y>=0
<=>
y^2+√(y+2006)=2006
√(y+2006)=z=>z>=√(2006)
=>hệ
y^2+z=2006
z^2-y=2006
trừ cho nhau
y^2-z^2+y+z=0
(y+z)(y-z+1)=0
=>
(y-z+1)=0 (*)
hoặc y+z=0 loại luôn
(*)=>√(y+2006)=y+1
=>y^2+y-2005=0=>y=(√8021-1)/2 nghiệm (-)loại
=>x=+-√[(√8021-1)/2]

Ư(297)={1;3;9;11;27;33;99;297}
Vậy * có thể là 3;9
Vậy ** có thể là: 99;33

(Cái này kết luận theo đề bài)

b) =>\(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\3-\dfrac{1}{2}x=0\\1+\dfrac{3}{4}:x=0\end{matrix}\right.=>\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}x=3\\\dfrac{3}{4x}=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\x=6\\x=\dfrac{-3}{4}\end{matrix}\right.\)