\(1.\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(2.\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}+1}{\sqrt{5}-1 }+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\)
\(3.\left(\dfrac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\dfrac{\sqrt{5}-\sqrt{3}}{2}=\left(\sqrt{3}-\sqrt{5}\right)\cdot\dfrac{2}{\sqrt{5}-\sqrt{3}}=-2\)
\(4.\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\dfrac{\sqrt{3}\left(\sqrt{3}+6\right)}{\sqrt{3}}-\dfrac{13\cdot\left(4-\sqrt{3}\right)}{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)}=-\sqrt{3}+\sqrt{3}+6-4+\sqrt{3}=\sqrt{3}+2\)
\(5.\dfrac{\sqrt{21}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}+\dfrac{4\cdot\left(5+\sqrt{21}\right)}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}=\sqrt{21}+5+\sqrt{21}=2\sqrt{21}+5\)
\(6.\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\left(\sqrt{\sqrt{3}+1}+1\right)\left(\sqrt{\sqrt{3}+1}-1\right)}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-1\right)}{\left(\sqrt{\sqrt{3}+1}-1\right)\left(\sqrt{\sqrt{3}+1}+1\right)}=\sqrt{\sqrt{3}+1}+1-\sqrt{\sqrt{3}+1}+1=2\)

Mai mình làm cho nha, giờ đi soạn văn đã...

\(1.Q=\dfrac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x+2}}\)
       \(=\dfrac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{\left(x-x\right)+\left(2\sqrt{x}-\sqrt{x}\right)-\left(10-4-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
       \(=\dfrac{1}{\sqrt{x}+2}\)
\(2.\)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:
\(\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{9}\)\(\left(x\ge0;x\ne9\right)\)
\(< =>\sqrt{x}+2=9\)
\(< =>x=49\left(TM\right)\)
 

B

\(3.\left(\sqrt{7}+3\right)\cdot\left(3-\sqrt{7}\right)=2\)
\(4.\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{6}{\sqrt{3}\cdot\left(\sqrt{3}-1\right)}=\sqrt{3}+3+\sqrt{3}=3+2\sqrt{3}\)

7,10