\(1,Q=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+2}\\ 2,x=16\Leftrightarrow Q=\dfrac{1}{4+2}=\dfrac{1}{6}\\ 3,Q=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\Leftrightarrow x=1\left(tm\right)\\ 4,\Leftrightarrow Q-\dfrac{1}{9}=\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{9}>0\\ \Leftrightarrow\dfrac{9-\sqrt{x}-2}{9\left(\sqrt{x}+2\right)}>0\\ \Leftrightarrow7-\sqrt{x}>0\left(\sqrt{x}+2>0\right)\\ \Leftrightarrow\sqrt{x}< 7\Leftrightarrow0\le x< 49;x\ne9\)
\(5,Q=\dfrac{1}{\sqrt{x}+2}\le\dfrac{1}{0+2}=\dfrac{1}{2}\\ Q_{max}=\dfrac{1}{2}\Leftrightarrow x=0\)
\(1.Q=\dfrac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x+2}}\)
\(=\dfrac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(x-x\right)+\left(2\sqrt{x}-\sqrt{x}\right)-\left(10-4-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
\(2.\)Với x=16(TM) thì:\(Q=\dfrac{1}{\sqrt{16}+2}=\dfrac{1}{6}\)
3. Để Q=\(\dfrac{1}{3}\) thì:
\(\dfrac{1}{\sqrt{x}+2}=\dfrac{1}{3}\left(x\ge0,x\ne9\right)\)
\(< =>\sqrt{x}+2=3\)
\(< =>x=1\left(TM\right)\)
Vậy với x=1 thì Q=\(\dfrac{1}{3}\)
4. Để \(Q\ge\dfrac{1}{9}\) thì:
\(\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{9}\)\(\left(x\ge0;x\ne9\right)\)
\(< =>\sqrt{x}+2=9\)
\(< =>x=49\left(TM\right)\)
