a, Thay x = 9 ta được
\(A=\dfrac{3+3}{3}=2\)
b, đk x >= 0 ; x khác 9
sửa đề \(B=\left(\dfrac{x+3+\sqrt{x}-3}{x-9}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(x-9\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
c, Ta có \(C=A.B=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
4, \(C-\dfrac{3}{2}>0\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\Leftrightarrow\dfrac{-\sqrt{x}+2}{2\sqrt{x}}>0\)
TH1 \(\left\{{}\begin{matrix}-\sqrt{x}+2>0\\2\sqrt{x}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 4\\x>0\end{matrix}\right.\Leftrightarrow0< x< 4\)
TH2 : \(\left\{{}\begin{matrix}-\sqrt{x}+2< 0\\2\sqrt{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>4\\x< 0\end{matrix}\right.\left(voli\right)\)
