Prosiłabym o wszystkie obliczenia z wyjaśnieniem.

[tex]\dfrac{1}{\sqrt{2} } +\dfrac{1}{\sqrt{8} } +\dfrac{1}{\sqrt{32} } =\dfrac{\sqrt{16} }{\sqrt{2}\cdot \sqrt{16} } +\dfrac{\sqrt{4} }{\sqrt{8}\cdot \sqrt{4} } +\dfrac{1}{\sqrt{32} } =\dfrac{4}{\sqrt{2\cdot 16} } +\dfrac{2}{\sqrt{8\cdot 4} } +\dfrac{1}{\sqrt{32} } =\dfrac{4}{\sqrt{32} } +\dfrac{2}{\sqrt{32} } +\dfrac{1}{\sqrt{32} } =\dfrac{4+2+1}{\sqrt{32} }=\dfrac{7}{\sqrt{32} }=\dfrac{7}{\sqrt{16\cdot 2} }=\dfrac{7}{4\sqrt{2} }\cdot \dfrac{\sqrt{2} }{\sqrt{2} } =\dfrac{7\sqrt{2} }{8 }[/tex]

[tex]Odp:~~D.~~\dfrac{7}{4\sqrt{2} }[/tex]

Basetlaviz Basetlaviz · Answer 2

W mianownikach mamy:

[tex]\sqrt{2}\\\\\sqrt{8} = \sqrt{4\cdot2} = \sqrt{4}\cdot\sqrt{2} = 2\sqrt{2}\\\\\sqrt{32} = \sqrt{16\cdot2} = \sqrt{16}\cdot\sqrt{2} = 4\sqrt{2}[/tex]

Zatem wspólnym mianownikiem będzie [tex]4\sqrt{2}[/tex]

[tex]\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{8}} + \frac{1}{\sqrt{32}} = \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}+\frac{1}{4\sqrt{2}} = \frac{4}{4\sqrt{2}}+\frac{2}{4\sqrt{2}}+\frac{1}{4\sqrt{2}}=\boxed{\frac{7}{4\sqrt{2}}}\\\\\underline{Odp. \ D.}[/tex]

Po usunięciu niewymierności z mianownika otrzymujemy:

[tex]\frac{7}{4\sqrt{2}} = \frac{7}{4\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{4\cdot2} = \frac{7\sqrt{2}}{8}[/tex]