Question

imi rezolvati si mie exercitiul 2 va rog pt ca nu imi iese si o am tema pe luni.Va multumesc.

Answer 1

√(7+√7) × √[√14+√(7+√7)]× √[√14-√(7+√7)]=

=√(7+√7) × √[(√14)²-(√(7+√7))²=

=√(7+√7) × √(14-7-√7)=
=√(7+√7) × √(7-√7)=
=√49-7=√42

Answer 2

   
[tex]2) \\ \sqrt{7+\sqrt{7} } \times \sqrt{\sqrt{14}+ \sqrt{7+\sqrt{7} }} \times \sqrt{\sqrt{14}- \sqrt{7+\sqrt{7} }} = \\ \\ =\sqrt{7+\sqrt{7} } \times \sqrt{\Big(\sqrt{14}+ \sqrt{7+\sqrt{7} }\Big) \times \Big(\sqrt{14}- \sqrt{7+\sqrt{7} }\Big)} = \\ \\ =\sqrt{7+\sqrt{7} } \times \sqrt{\Big(\sqrt{14}\Big)^2- \Big(\sqrt{7+\sqrt{7} }\Big)^2 } = \\ \\ =\sqrt{7+\sqrt{7} } \times \sqrt{14- (7+\sqrt{7}) } } = \\ \\ =\sqrt{7+\sqrt{7} } \times \sqrt{14- 7-\sqrt{7} } }=[/tex]


[tex]=\sqrt{7+\sqrt{7} } \times \sqrt{7-\sqrt{7} } }= \\ \\ =\sqrt{(7+\sqrt{7}) \times (7-\sqrt{7} }) }= \\ \\ =\sqrt{7^2-(\sqrt{7})^2}=\sqrt{49-7}= \boxed{\sqrt{42} } [/tex]