Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ 3\sqrt{3} ;\ b.\ \sqrt{5} +\sqrt{2} ;\ c.\ \left(\sqrt{2} +1\right)^{2}\\ d.\frac{\left(\sqrt{5} -\sqrt{3}\right)^{2}}{2} ;\ e.\ 1+\sqrt{a} +a\\ f.\ \frac{\sqrt{18} +\sqrt{8} +2\sqrt{2}}{\left(\sqrt{18} +\sqrt{8}\right)^{2} -8} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{9}{\sqrt{3}} =\frac{9\sqrt{3}}{3} =3\sqrt{3}\\ b.\ \frac{3}{\sqrt{5} -\sqrt{2}} =\frac{3\left(\sqrt{5} +\sqrt{2}\right)}{5-2} =\frac{3\left(\sqrt{5} +\sqrt{2}\right)}{3} =\sqrt{5} +\sqrt{2}\\ c.\ \frac{\sqrt{2} +1}{\sqrt{2} -1} =\frac{\left(\sqrt{2} +1\right)^{2}}{2-1} =\left(\sqrt{2} +1\right)^{2}\\ d.\ \frac{\sqrt{5} -\sqrt{3}}{\sqrt{5} +\sqrt{3}} =\frac{\left(\sqrt{5} -\sqrt{3}\right)^{2}}{5-3} =\frac{\left(\sqrt{5} -\sqrt{3}\right)^{2}}{2}\\ e.\ \frac{1-a\sqrt{a}}{1-\sqrt{a}} =\frac{\left( 1-\sqrt{a}\right)\left( 1+\sqrt{a} +a\right)}{1-\sqrt{a}} =1+\sqrt{a} +a\\ f.\ \frac{1}{\sqrt{18} +\sqrt{8} -2\sqrt{2}} =\frac{\sqrt{18} +\sqrt{8} +2\sqrt{2}}{\left(\sqrt{18} +\sqrt{8}\right)^{2} -\left( 2\sqrt{2}\right)^{2}}\\ =\frac{\sqrt{18} +\sqrt{8} +2\sqrt{2}}{\left(\sqrt{18} +\sqrt{8}\right)^{2} -8} \end{array}$
