[tex]zad.1\\\\\dfrac{\sqrt{50} -\sqrt{18} }{\sqrt{2} } =\dfrac{\sqrt{25\cdot 2} -\sqrt{9\cdot 2} }{\sqrt{2} } =\dfrac{\sqrt{5^{2} \cdot 2} -\sqrt{3^{2} \cdot 2} }{\sqrt{2} } =\dfrac{5\sqrt{2} -3\sqrt{2} }{\sqrt{2} } =\dfrac{2\sqrt{2} }{\sqrt{2} } =2[/tex]
[tex]zad.2\\\\\dfrac{\sqrt[3]{2} }{\sqrt[3]{4} } \cdot \dfrac{\sqrt[3]{16} }{\sqrt[3]{16} }=\dfrac{\sqrt[3]{2 \cdot 16} }{\sqrt[3]{4\cdot 16} }=\dfrac{\sqrt[3]{32} }{\sqrt[3]{4\cdot \cdot 4\cdot 4} }=\dfrac{\sqrt[3]{8\cdot 4} }{\sqrt[3]{4^{3} } }=\dfrac{\sqrt[3]{2^{3} \cdot 4} }{4 }=\dfrac{2\sqrt[3]{4} }{4 }=\dfrac{\sqrt[3]{4} }{2 }[/tex]
[tex]zad.3\\\\\dfrac{(3\sqrt{2} -6\sqrt{7} )^{0} \cdot 2}{4} =\dfrac{1\cdot 2}{4} =\dfrac{2}{4} =\dfrac{1}{2}[/tex]
Pamietamy : Każda liczba niezerowa podniesiona do potęgi zerowej równa się 1.
[tex]zad.4\\\\\\\\ \sqrt{\dfrac{9}{7} } +\sqrt{\dfrac{16}{7} } =\dfrac{\sqrt{9} }{\sqrt{7} } +\dfrac{\sqrt{16} }{\sqrt{7} }=\dfrac{\sqrt{3^{2} } }{\sqrt{7} }+\dfrac{\sqrt{4^{2} } }{\sqrt{7} }=\dfrac{3}{\sqrt{7} } +\dfrac{4}{\sqrt{7} } =\dfrac{3+4}{\sqrt{7} } =\dfrac{7}{\sqrt{7} } \cdot \dfrac{\sqrt{7} }{\sqrt{7} } =\dfrac{7\sqrt{7} }{7} =\sqrt{7}[/tex]
