[tex]\displaystyle\\
\frac{12}{\sqrt{7}-\sqrt{3}}-\frac{14}{\sqrt{7}}+\frac{6}{\sqrt{3}}=\\\\
\text{Rationalizam numitorii.}\\\\
=\frac{12(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}-\frac{14\sqrt{7}}{\sqrt{7}\times\sqrt{7}}+\frac{6\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\\\\
=\frac{12(\sqrt{7}+\sqrt{3})}{7-3}-\frac{14\sqrt{7} }{7}+\frac{6\sqrt{3}}{3}=\\\\
=\frac{12(\sqrt{7}+ \sqrt{3})}{4}-\frac{14\sqrt{7}}{7}+\frac{6\sqrt{3}}{3}=\\\\
\text{Aducem fractiile la acelasi numitor care este 84.}[/tex]
[tex]\displaystyle\\
=\frac{21\times12(\sqrt{7}+ \sqrt{3})}{21\times 4}-\frac{12\times14\sqrt{7}}{12\times7}+\frac{28\times6\sqrt{3}}{28\times3}=\\\\
=\frac{252(\sqrt{7}+ \sqrt{3})}{84}-\frac{168\sqrt{7}}{84}+\frac{168\sqrt{3}}{84}=\\\\
=\frac{252(\sqrt{7}+\sqrt{3}) -168\sqrt{7} +168\sqrt{3}}{84}=\\\\
=\frac{252(\sqrt{7}+ \sqrt{3})}{84}=\boxed{\bf 3(\sqrt{7}+ \sqrt{3})}[/tex]
