Question

Calculand $\frac{12}{√7-√3}$-$\frac{14}{√7}$+$\frac{6}{√3}$ se obtine...
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Answer 1

   
[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]