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Trả lời:
ĐKXĐ: $x\ge 0;x\ne 9$
$a,$
$\begin{array}{l}M=\bigg{(}\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\bigg{)}:\bigg{(}\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\bigg{)}\\\,\,\,\,\,\,\,\,=\bigg{[}\dfrac{(\sqrt{x}+3)(\sqrt{x}+3)}{x-9}-\dfrac{(\sqrt{x}-3)(\sqrt{x}-3)}{x-9}\bigg{]}:\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\\\,\,\,\,\,\,\,\,=\dfrac{x+6\sqrt{x}+9-x+6\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}+3}{-3}\\\,\,\,\,\,\,\,\,=\dfrac{12\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}+3}{-3}\\\,\,\,\,\,\,\,\,=\dfrac{-4\sqrt{x}}{\sqrt{x}-3}\end{array}$
$b,$ $M>0$
$⇔\dfrac{-4\sqrt{x}}{\sqrt{x}-3}>0$
Ta có: $-4\sqrt{x}\ge 0$
$⇒\sqrt{x}-3<0$
$⇒\sqrt{x}<3$
$⇒0\le x\le 9$
Vậy $M$ có giá trị dương khi $0\le x\le 9$.
