Proszę o szybką pomoc daje naj
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I.
[tex]a) \frac1{\sqrt7} = \frac1{\sqrt7} * \frac{\sqrt7}{\sqrt7}=\frac{\sqrt7}7\\b) \frac1{4\sqrt5} = \frac1{4\sqrt5} * \frac{\sqrt5}{\sqrt5}=\frac{\sqrt5}{4*5}=\frac{\sqrt5}{20}\\c) \frac{6\sqrt2}{\sqrt3}=\frac{6\sqrt2}{\sqrt3} * \frac{\sqrt3}{\sqrt3}=\frac{6\sqrt6}{3}=2\sqrt6\\[/tex]
II.
[tex]\frac{3\sqrt2}{2\sqrt8}=\frac{3\sqrt2}{2\sqrt{4*2}}=\frac{3\sqrt2}{{4\sqrt2}}=\frac34[/tex]
Odp. B
Odpowiedź i szczegółowe wyjaśnienie:
[tex]1.\\a)\ \dfrac{1}{\sqrt7}=\dfrac{1}{\sqrt7}\cdot\dfrac{\sqrt7}{\sqrt7}=\dfrac{\sqrt7}{7}\\\\\\b)\ \dfrac{1}{4\sqrt5}=\dfrac{1}{4\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{\sqrt{5}}{4\cdot5}=\dfrac{\sqrt{5}}{20}\\\\\\c)\ \dfrac{6\sqrt2}{\sqrt3}=\dfrac{6\sqrt2}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{6\sqrt6}{3}=2\sqrt6[/tex]
[tex]2.\ \dfrac{3\sqrt2}{2\sqrt8}=\dfrac{3\sqrt2}{2\cdot\sqrt{4\cdot2}}=\dfrac{3\sqrt2}{2\cdot\sqrt4\cdot\sqrt2}=\dfrac{3\sqrt2}{2\cdot2\cdot\sqrt2}=\dfrac{3}{2\cdot2}=\dfrac34[/tex]