Bài 3: 

a) Ta có: \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}-\dfrac{a-4}{\sqrt{a}-2}\)

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

=0

b) Ta có: \(\dfrac{9-a}{\sqrt{a}+3}-\dfrac{a-6\sqrt{a}+9}{\sqrt{a}-3}\)

\(=3-\sqrt{a}-\sqrt{a}+3\)

\(=6-2\sqrt{a}\)

c) Ta có: \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)\)

=0

d) Ta có: \(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\)

\(=a+\sqrt{ab}+b+\sqrt{ab}\)

\(=a+2\sqrt{ab}+b\)

Bài 1:

a.

\(\frac{4\sqrt{5}+\sqrt{15}}{\sqrt{5}}=\frac{\sqrt{5}(4+\sqrt{3})}{\sqrt{5}}=4+\sqrt{3}\)

$\frac{7-\sqrt{7}}{3\sqrt{7}}=\frac{\sqrt{7}(\sqrt{7}-1)}{3\sqrt{7}}=\frac{\sqrt{7}-1}{3}$
\(\frac{4\sqrt{2}-\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}(4-\sqrt{3})}{\sqrt{2}.\sqrt{6}}=\frac{4-\sqrt{3}}{\sqrt{6}}\)

\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\frac{(3\sqrt{2}-2\sqrt{3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}=\frac{\sqrt{6}}{3-2}=\sqrt{6}\)

b.

\(\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}(\sqrt{a}-2)}{\sqrt{a}-2}=\sqrt{a}\)

\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}=\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}=1+\sqrt{a}+a\)

\(\frac{a+10\sqrt{a}+25}{\sqrt{a}+5}=\frac{(\sqrt{a}+5)^2}{\sqrt{a}+5}=\sqrt{a}+5\)

\(\frac{a-9}{\sqrt{a}+3}=\frac{(\sqrt{a}-3)(\sqrt{a}+3)}{\sqrt{a}-3}=\sqrt{a}+3\)

Bài 2.

a.

\(\frac{6-\sqrt{6}}{\sqrt{6}-1}+\frac{6+\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}(\sqrt{6}-1)}{\sqrt{6}-1}+\frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{6}}=\sqrt{6}+(\sqrt{6}+1)=2\sqrt{6}+1\)

b.

\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}+\frac{-\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)

c.

\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}(\sqrt{5}-2)}{\sqrt{5}-2}+\frac{\sqrt{6}(\sqrt{6}+2)}{\sqrt{3}+\sqrt{2}}\)

\(=\sqrt{3}+\frac{\sqrt{6}.\sqrt{2}(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{12}=3\sqrt{3}\)

Bài 2:

d.

\(\frac{3\sqrt{2}-6}{\sqrt{2}-1}+\frac{6\sqrt{2}-4}{\sqrt{2}-3}=\frac{3\sqrt{2}(1-\sqrt{2})}{\sqrt{2}-1}+\frac{2\sqrt{2}(3-\sqrt{2})}{\sqrt{2}-3}\)

\(=-3\sqrt{2}-2\sqrt{2}=-5\sqrt{2}\)

e.

$\frac{1}{1+\sqrt{5}}+\frac{1}{1-\sqrt{5}}=\frac{1-\sqrt{5}+1+\sqrt{5}}{(1+\sqrt{5})(1-\sqrt{5})}=\frac{2}{1-5}=\frac{2}{-4}=\frac{-1}{2}$
f.

$\frac{1}{3-2\sqrt{3}}+\frac{1}{3+2\sqrt{3}}=\frac{3+2\sqrt{3}+3-2\sqrt{3}}{(3-2\sqrt{3})(3+2\sqrt{3})}=\frac{6}{9-12}=-2$

Bài 2.

g.

$\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{2}{1-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}}{(\sqrt{5}+\sqrt{7})(\sqrt{5}-\sqrt{7})}+\frac{2(1+\sqrt{7})}{(1-\sqrt{7})(1+\sqrt{7})}$

$=\frac{\sqrt{5}-\sqrt{7}}{-2}+\frac{2(1+\sqrt{7})}{1-7}$

$=\frac{\sqrt{7}-2-3\sqrt{5}}{6}$

h.

$=\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}-\frac{(3-\sqrt{2})^2}{(3-\sqrt{2})(3+\sqrt{2})}$

$=\frac{3-2\sqrt{2}}{2-1}-\frac{11-6\sqrt{2}}{9-2}=3-2\sqrt{2}-\frac{11-6\sqrt{2}}{7}=\frac{10-8\sqrt{2}}{7}$

Bài 3.

a.

\(=\frac{(\sqrt{a}+2)^2}{\sqrt{a}+2}-\frac{(\sqrt{a}-2)(\sqrt{a}+2)}{\sqrt{a}-2}=\sqrt{a}+2-(\sqrt{a}+2)=0\)

b.

\(=\frac{(3-\sqrt{a})(3+\sqrt{a})}{\sqrt{a}+3}-\frac{(\sqrt{a}-3)^2}{\sqrt{a}-3}=3-\sqrt{a}-(\sqrt{a}-3)=2(3-\sqrt{a})\)

c.

\(\frac{(\sqrt{a}-\sqrt{b})^2}{\sqrt{a}-\sqrt{b}}-\frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}-(\sqrt{a}-\sqrt{b})=0\)

d.

\(=\frac{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}=a+\sqrt{ab}+b+\sqrt{ab}=a+2\sqrt{ab}+b=(\sqrt{a}+\sqrt{b})^2\)