Question

exercitiul 3 de la sub.II
va rog

Answer 1

Rezolvarea in atașament

Answer 2

Răspuns:

$a=|\sqrt{5}-\sqrt{7}|+\dfrac{2}{\sqrt{7}-\sqrt{5}}-\sqrt{28}=|\sqrt{7}-\sqrt{5}|+\dfrac{2(\sqrt{7}+\sqrt{5})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}-\sqrt{4*7}=\\=\sqrt{7}-\sqrt{5}+\dfrac{2(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}-2\sqrt{7}=\sqrt{7}-\sqrt{5}+\dfrac{2(\sqrt{7}+\sqrt{5})}{7-5}-2\sqrt{7}=\\=\sqrt{7}-\sqrt{5}+\sqrt{7}+\sqrt{5}-2\sqrt{7}=2\sqrt{7}-2\sqrt{7}=0$Deci a∈Z

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