Ta có :

a.  $\sqrt{18(\sqrt{2}-\sqrt{3})^{2}}$

=  $\sqrt{3^{2}.2.(\sqrt{2}-\sqrt{3})^{2}}$

= $3\left | \sqrt{2}-\sqrt{3} \right |\sqrt{2}=3(\sqrt{2}-\sqrt{3})\sqrt{2}$

= $3\sqrt{6}-6$

Vậy $\sqrt{18(\sqrt{2}-\sqrt{3})^{2}}=3\sqrt{6}-6$

b.  $ab\sqrt{1+\frac{1}{a^{2}b^{2}}}$

= $ab\sqrt{\frac{a^{2}b^{2}+1}{a^{2}b^{2}}}$

= $\frac{ab\sqrt{a^{2}b^{2}+1}}{\left | ab \right |}$

= $\left\{\begin{matrix}\sqrt{a^{2}b^{2}+1}(ab>0) & \\ -\sqrt{a^{2}b^{2}+1}(ab<0)& \end{matrix}\right.$

Vậy  $ab\sqrt{1+\frac{1}{a^{2}b^{2}}}$ = $\left\{\begin{matrix}\sqrt{a^{2}b^{2}+1}(ab>0) & \\ -\sqrt{a^{2}b^{2}+1}(ab<0)& \end{matrix}\right.$

c.   $\sqrt{\frac{a}{b^{3}}+\frac{a}{b^{4}}}$ 

= $\sqrt{\frac{ab+a}{b^{4}}}$

= $\frac{\sqrt{ab +a}}{\left | b^{2} \right |}=\frac{\sqrt{ab+a}}{b^{2}}$

Vậy $\sqrt{\frac{a}{b^{3}}+\frac{a}{b^{4}}}$  = $\frac{\sqrt{ab+a}}{b^{2}}$

d.   $\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}$

= $\frac{\sqrt{a}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}=\sqrt{a}$

Vậy  $\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\sqrt{a}$