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Đáp án:

a) $Q=\dfrac{1}{\sqrt{x}-3}$

b) $x=36$ thì $Q=\dfrac{1}{3}$

c) $9<x<16$ thì $Q>1$

Giải thích các bước giải:

a)

$Q=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{3}{x\sqrt{x}-9\sqrt{x}}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3\sqrt{x}-3}{x+3\sqrt{x}}\right)\,\,\,(x>0;x\ne9)\\=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}(\sqrt{x}-3)(\sqrt{x}+3)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3\sqrt{x}-3}{\sqrt{x}(\sqrt{x}+3)}\right)\\=\dfrac{\sqrt{x}(\sqrt{x}-3)+3}{\sqrt{x}(\sqrt{x}-3)(\sqrt{x}+3)}:\dfrac{x-3\sqrt{x}+3}{\sqrt{x}(\sqrt{x}+3)}\\=\dfrac{x-3\sqrt{x}+3}{\sqrt{x}(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}(\sqrt{x}+3)}{x-3\sqrt{x}+3}\\=\dfrac{1}{\sqrt{x}-3}$

b)

$Q=\dfrac{1}{\sqrt{x}-3}\,\,\,(x>0;x\ne9)$

Để $Q=\dfrac{1}{3}$

$\to\dfrac{1}{\sqrt{x}-3}=\dfrac{1}{3}\\\to\sqrt{x}-3=3\\⇔\sqrt{x}=6\\⇔x=36$ (thoả mãn)

Vậy $x=36$ thì $Q=\dfrac{1}{3}$

c)

$Q=\dfrac{1}{\sqrt{x}-3}\,\,\,(x>0;x\ne9)$

Để $Q>1$

$\to\dfrac{1}{\sqrt{x}-3}>1\\⇔\dfrac{1}{\sqrt{x}-3}-1>0\\⇔\dfrac{1-\sqrt{x}+3}{\sqrt{x}-3}>0\\⇔\dfrac{4-\sqrt{x}}{\sqrt{x}-3}>0\\⇔\left[\begin{array}{l}\begin{cases}4-\sqrt{x}>0\\\sqrt{x}-3>0\end{cases}\\\begin{cases}4-\sqrt{x}<0\\\sqrt{x}-3<0\end{cases}\end{array}\right.\\⇔\left[\begin{array}{l}\begin{cases}x<16\\x>9\end{cases}\\\begin{cases}x>16\\x<9\end{cases}\text{ (loai)}\end{array}\right.\\⇔9<x<16$

Kết hợp với điều kiện $x>0;x\ne9$

$\to 9<x<16$

Vậy $9<x<16$ thì $Q>1$