ĐKXĐ: \(x\ge0;a\ne4\)

\(H=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

\(=\frac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)

\(H< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}< 2\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}-2< 0\)

\(\Rightarrow\frac{\sqrt{a}-4-2\left(\sqrt{a}-2\right)}{\sqrt{a}-2}< 0\Rightarrow\frac{-\sqrt{a}}{\sqrt{a}-2}< 0\)

\(\Rightarrow\frac{\sqrt{a}}{\sqrt{a}-2}>0\Rightarrow\sqrt{a}-2>0\Rightarrow a>4\)

\(a^2+3a=0\Rightarrow a\left(a+3\right)=0\Rightarrow a=0\) (do \(a\ge0\Rightarrow a+3>0\))

\(\Rightarrow H=\frac{0-4}{0-2}=2\)

\(H=5\Rightarrow\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\Rightarrow\sqrt{a}-4=5\sqrt{a}-10\)

\(\Rightarrow4\sqrt{a}=6\Rightarrow\sqrt{a}=\frac{3}{2}\Rightarrow a=\frac{9}{4}\)

a: \(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)

\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

b: Để H<2 thì H-2<0

\(\Leftrightarrow\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}< 0\)

=>căn a-2>0

hay a>4

d: Để H=5 thì căn a-4=5 căn a-10

=>-4 căn a=-6

=>căn a=3/2

hay a=9/4

\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

a,Đk: a≥0 ; a khác 4

H=\(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\) -\(\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\) -\(\dfrac{1}{\sqrt{a}-2}\)

= \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

=\(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

=\(\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)

=\(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

=\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

b, Để H<2

<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) <2

<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -2<0

<=>\(\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}\) <0

<=>\(\dfrac{-\sqrt{a}}{\sqrt{a}-2}\) <0

<=>\(\left\{{}\begin{matrix}-\sqrt{a}< 0\\\sqrt{a}-2>0\end{matrix}\right.\) ( vì \(\sqrt{a}>0< =>-\sqrt{a}< 0\)

<=> a>4

vậy để H <2 khi a>4

c, Ta có a\(^2\) +3a=0

<=> a(a+3)=0

<=>a=0 hoặc a=-3(vô lí)

+ Với a=0 <=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =\(\dfrac{0-4}{0-2}\) =2

d, Để H=5

<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =5

<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -5=0

<=>\(\dfrac{\sqrt{a}-4-5\sqrt{a}+10}{\sqrt{a}-2}\) =0

<=>-4\(\sqrt{a}\) +6=0

<=> a=\(\dfrac{9}{4}\)

a) \(A=\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)

b) \(B=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)

c) \(C=\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{2\sqrt{2}+\sqrt{6}}{4+2\sqrt{3}}=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(4-2\sqrt{3}\right)}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\)

d) \(D=\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=-\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)\)