Question

Care ma ajuta si pe mine.Sunt la radicali compuse si profesoara noastra ne-a dat direct exercitii de 3 stele ,respectiv 4 stele.

Answer 1

$a)a= \sqrt{6-2 \sqrt{4+2 \sqrt{3} } } - \sqrt{8-2 \sqrt{6+2 \sqrt{5} } } = \\ = \sqrt{6-2 \sqrt{(1+ \sqrt{3} )^2} }- \sqrt{8-2 \sqrt{(1+ \sqrt{5} )^2} }= \sqrt{6-2(1+ \sqrt{3}) }- \\ - \sqrt{8-2(1+ \sqrt{5}) }= \sqrt{4-2 \sqrt{3} }- \sqrt{6-2 \sqrt{5} }= \sqrt{( \sqrt{3}-1 })^2- \\ - \sqrt{ (\sqrt{5}-1)^2 }= \sqrt{3}-1- \sqrt{5}+1= \sqrt{3}- \sqrt{5} \\ a^{2}=3+5-2 \sqrt{15}=8-2 \sqrt{15}$
$b)a= \sqrt{6-2 \sqrt{3}-2 \sqrt{( \sqrt{5}- \sqrt{3})^2 } }- \sqrt{8+2 \sqrt{3}+2 \sqrt{( \sqrt{7}- \sqrt{3} })^2 } = \\ = \sqrt{6-2 \sqrt{3}-2( \sqrt{5}- \sqrt{3)} }- \sqrt{8+2 \sqrt{3}+2( \sqrt{7}- \sqrt{3}) }= \\ = \sqrt{6-2 \sqrt{5} }- \sqrt{8+2 \sqrt{7} } = \sqrt{( \sqrt{5}-1)^2 }- \sqrt{(1+ \sqrt{7})^2 } = \\ = \sqrt{5}-1-1- \sqrt{7}$$= \sqrt{5}- \sqrt{7}-2$