Odpowiedź i szczegółowe wyjaśnienie:
[tex]\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt8}+\dfrac{1}{\sqrt{32}}=\dfrac{1}{\sqrt2}+\dfrac{1}{2\sqrt2}+\dfrac{1}{4\sqrt2}=\dfrac{4}{4\sqrt2}+\dfrac{2}{4\sqrt2}+\dfrac{1}{4\sqrt2}=\\\\=\dfrac{7}{4\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=\dfrac{7\sqrt2}{4\cdot2}=\dfrac{7\sqrt2}{8}[/tex]
