1) `M=\frac{ \sqrt{x}}{ \sqrt{x}+3}+ \frac{ 2\sqrt{x}}{ \sqrt{x}-3}- \frac{3x+9}{x-9}`
ĐKXĐ:` x≥0`;`xne9`
$=\dfrac{ \sqrt{x}(\sqrt{x}-3)}{ (\sqrt{x}+3)(\sqrt{x}-3)}+ \dfrac{ 2\sqrt{x}(\sqrt{x}+3)}{ (\sqrt{x}-3)(\sqrt{x}+3)}- \dfrac{3x+9}{x-9}$
$= \dfrac{\sqrt{x}(\sqrt{x}-3)+ 2\sqrt{x}(\sqrt{x}+3)-3x-9}{(\sqrt{x}+3)(\sqrt{x}-3)}$
$= \dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{(\sqrt{x}+3)(\sqrt{x}-3)}$
$= \dfrac{3\sqrt{x}-9}{(\sqrt{x}+3)(\sqrt{x}-3)}$
$= \dfrac{3(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}$
$= \dfrac{3 }{\sqrt{x}+3 }$
2) $M=\dfrac{1 }{3 }$ ĐKXĐ:` x≥0`;`xne9`
$⇒ \dfrac{3 }{\sqrt{x}+3 }=\dfrac{1 }{3 }$
$⇔\dfrac{9 }{3(\sqrt{x}+3) }=\dfrac{ \sqrt{x}+3}{3(\sqrt{x}+3) }$
$⇒\sqrt{x}+3=9$
$⇔\sqrt{x}=6$
`⇔x=36`
Vậy `S={36}`
3) Để M là số nguyên thì `3⋮ \sqrt{x}+3` ĐKXĐ:` x≥0`;`xne9`
`⇔\sqrt{x}+3` là ước của `3`
⇒\(\left[ \begin{array}{l}\sqrt{x}+3=1\\\sqrt{x}+3=-1 \\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\sqrt{x}=1-3\\\sqrt{x}=-1-3 \\\sqrt{x}=3-3\\\sqrt{x}=-3-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}\sqrt{x}=-2\\\sqrt{x}=-4 \\\sqrt{x}= 0\\\sqrt{x}=-6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\\ x=16 \\x= 0\\x=36\end{array} \right.\)
Vậy `S={4;16;0;36}`
