Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\sqrt{6} +\sqrt{7}\\ b.3\sqrt{2} -2\sqrt{3}\\ c.\frac{4-\sqrt{6}}{10}\\ d.\frac{-6}{\sqrt{2}} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\frac{3}{\sqrt{6} -\sqrt{3}} +\frac{4}{\sqrt{7} +\sqrt{3}}\\ =\frac{3\left(\sqrt{6} +\sqrt{3}\right)}{( 6-3)} +\frac{4\left(\sqrt{7} -\sqrt{3}\right)}{7-3}\\ =\sqrt{6} +\sqrt{3} +\sqrt{7} -\sqrt{3} =\sqrt{6} +\sqrt{7}\\ b.\ \frac{6}{3\sqrt{2} +2\sqrt{3}} =\frac{6\left( 3\sqrt{2} -2\sqrt{3}\right)}{\left( 3\sqrt{2}\right)^{2} -\left( 2\sqrt{3}\right)^{2}}\\ =\frac{6\left( 3\sqrt{2} -2\sqrt{3}\right)}{6} =3\sqrt{2} -2\sqrt{3}\\ c.\ \frac{\sqrt{2}}{2\sqrt{3} +4\sqrt{2}} =\frac{1}{\sqrt{6} +4}\\ =\frac{\sqrt{6} -4}{\left(\sqrt{6}\right)^{2} -4^{2}} =\frac{\sqrt{6} -4}{-10} =\frac{4-\sqrt{6}}{10}\\ d.\ \frac{1}{4-3\sqrt{2}} -\frac{1}{4+3\sqrt{2}}\\ =\frac{4+3\sqrt{2} -\left( 4-3\sqrt{2}\right)}{\left( 4+3\sqrt{2}\right) .\left( 4-3\sqrt{2}\right)} =\frac{6\sqrt{2}}{-2} =\frac{-6}{\sqrt{2}} \end{array}$
