Răspuns:
[tex](\sqrt{\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}} +\sqrt{\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}}} +\sqrt{\frac{\sqrt{3}-1}{\sqrt{3}+1}}} ):\frac{2\sqrt{2}}{\sqrt{14}+\sqrt{2}} =[/tex]
[tex]=(\sqrt{\frac{\sqrt{7}-\sqrt{5})\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}} +\sqrt{\frac{\sqrt{5}-\sqrt{3})\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}}} +\sqrt{\frac{\sqrt{3}-1)\sqrt{3}-1}{\sqrt{3}+1}}} ):\frac{2\sqrt{2}}{\sqrt{2}(\sqrt{7}+1)} =[/tex]
[tex]=(\sqrt{\frac{(\sqrt{7}-\sqrt{5})^{2} }{7-5}}+\sqrt{\frac{(\sqrt{5}-\sqrt{3})^{2} }{5-3}}+\sqrt{\frac{(\sqrt{3}-1)^{2} }{3-1}})*\frac{\sqrt{7}+1}{2}=[/tex]
[tex]=\frac{\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{\sqrt{2}} *\frac{\sqrt{7}+1}{2} =[/tex]
[tex]=\frac{(\sqrt{7}-1)(\sqrt{7}+1)}{2\sqrt{2}} =\frac{\sqrt{2})7-1}{2\sqrt{2}} =\frac{6\sqrt{2}}{4}=\frac{3\sqrt{2}}{2}[/tex]
Explicație pas cu pas:
Pentru rezolvarea acestui exercitiu se foloseste formula de calcul prescurtat:
(a+b)(a-b)=a²-b²
