Răspuns :
[tex]a= \frac{1}{2-\sqrt{3}}\\ a= \frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}\\ a=2+ \sqrt{3}\\
b= \frac{1}{2+\sqrt{3}}\\
b=\frac{2-\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}\\
b=2- \sqrt{3}\\
(a+b)^2=(2+\sqrt{3}+2-\sqrt{3})^2=4^2=16 [/tex]
[tex](( \frac{1}{2}- \sqrt{3})+( \frac{1}{2}+ \sqrt{3} ))^{2}=( \frac{1}{2}- \sqrt{3})^{2}+2*( \frac{1}{2}- \sqrt{3})( \frac{1}{2}+ \sqrt{3})+\\+( \frac{1}{2}+ \sqrt{3})^{2}=(\frac{1}{2})^{2}- 2* \frac{1}{2}* \sqrt{3} + (\sqrt{3})^{2} +2((\frac{1}{2})^{2}-(\sqrt{3})^{2})+(\frac{1}{2})^{2}+ 2* \frac{1}{2}* \sqrt{3}+\\(\sqrt{3})^{2}= \frac{1}{4}- \sqrt{3} +3+2* \frac{1}{4}-6+ \frac{1}{4}+ \sqrt{3}+3= \frac{1}{4}+ \frac{1}{2} + \frac{1}{4}= \\=\frac{2}{4}+\frac{1}{2}=\frac{2+2}{4} =\frac{4}{4}=1[/tex]
