Question

Rationalizati numitorii fractiilor: a) 1 supra 2+ radical din 3; b) 5 supra radical din 7- radical din 2; c) 2 supra 3 radical din 2 + 2 radical din 3.

Answer 1

   
[tex]\displaystyle \texttt{Aplicam formula: } ~~ (a+b)(a-b)=a^2-b^2\\ \\ a)~~~ \frac{1}{2+ \sqrt{3} } = \frac{1\times (2-\sqrt{3})}{(2+ \sqrt{3})(2-\sqrt{3})}= \frac{2-\sqrt{3}}{4-3} = \boxed{2-\sqrt{3}} \\ \\ b)~~~ \frac{5}{\sqrt{7}-\sqrt{2}} =\frac{5(\sqrt{7}+\sqrt{2})}{(\sqrt{7}-\sqrt{2})(\sqrt{7}+\sqrt{2})} = \frac{5(\sqrt{7}+\sqrt{2})}{7-2} =\boxed{\sqrt{7}+\sqrt{2}} [/tex]

[tex]\displaystyle \\ \\ c)~~~ \frac{2}{3 \sqrt{2}+2 \sqrt{3}}= \frac{2(3 \sqrt{2}-2 \sqrt{3})}{(3 \sqrt{2}+2 \sqrt{3})(3 \sqrt{2}-2 \sqrt{3})}= \\ \\ = \frac{2(3 \sqrt{2}-2 \sqrt{3})}{9\times2 - 4 \times 3} = \frac{2(3 \sqrt{2}-2 \sqrt{3})}{18 - 12} = \frac{2(3 \sqrt{2}-2 \sqrt{3})}{6} = \boxed{\frac{3 \sqrt{2}-2 \sqrt{3}}{3} } [/tex]