2)[tex] (\sqrt{3- \sqrt{5} } + \sqrt{3+ \sqrt{5} } ) ^{2} =( \sqrt{3} -5+ \sqrt{3} +5)^2=
\sqrt{3} ^2-5^2+ \sqrt{3} ^2+5^2[/tex]=3-25+3+25 reducem -25 si 25 si ne ramane 3+3=6
3)a) ma=[tex] \frac{2- \sqrt{3}+2+ \sqrt{3} }{2} reducem cei doi radicali asemea cu semn diferit si ne ramane [/tex][tex] \frac{2+2}{2} = \frac{4}{2} =2[/tex]
mg=[tex] \sqrt{(2+ \sqrt{3})*(2- \sqrt{3} ) } = \sqrt{2^2* \sqrt{3}^2 } = \sqrt{4*3} = \sqrt{12} =3,4[/tex]
b)ma=[tex] \frac{5+ \sqrt{11}+5- \sqrt{11} }{2} reducem cei doi radicali asemenea dar cu semn diferit
= \frac{5+5}{2}[/tex]=[tex] \frac{10}{2} =5[/tex]
mg=[tex] \sqrt{(5+ \sqrt{11})*(5- \sqrt{11}) } = \sqrt{5^2- \sqrt{11} ^2} = \sqrt{25-11} = \sqrt{14} =3,7[/tex]
c)ma=[tex] \frac{( \sqrt{14} + \sqrt{10}^2)+(\sqrt{14} - \sqrt{10}^2)}{2} = \frac{14+10+14-10}{2} reducem termenii asemenea cu semn diferit[/tex]=[tex] \frac{14+14}{2} = \frac{28}{2} =14[/tex]
a=([tex]( \sqrt{14}+ \sqrt{10} )^2=14+10=24[/tex]
b=([tex]( \sqrt{14}- \sqrt{10} )^2=14-10=4[/tex]
mg=[tex] \sqrt{24*4} = \sqrt{96} =9,7[/tex]
