Odpowiedź :
zad.1
[tex]a)\\\\\sqrt[3]{28} \div \sqrt[3]{\dfrac{7}{2} }= \sqrt[3]{2^{3} } \\\\L=\sqrt[3]{28} \div \sqrt[3]{\dfrac{7}{2} }= \sqrt[3]{28\div \frac{7}{2} } =\sqrt[3]{28\cdot \frac{2}{7} } =\sqrt[3]{8} =\sqrt[3]{2^{3} } \\\\P =\sqrt[3]{2^{3} } \\\\L=P \\\\Odp: P-zapis~~prwadziwy.[/tex]
[tex]c)\\\\\sqrt[3]{0,75\div \dfrac{2}{9} } =\dfrac{3}{2} \\\\L=\sqrt[3]{0,75\div \dfrac{2}{9} } =\sqrt[3]{\dfrac{3}{4} \div \dfrac{2}{9} } =\sqrt[3]{\dfrac{3}{4} \cdot \dfrac{9}{2} } =\sqrt[3]{\dfrac{27}{8} }=\sqrt[3]{\dfrac{3^{3} }{2^{3} } } =\dfrac{3}{2} \\\\P=\dfrac{3}{2}\\\\L=P\\\\Odp:P-zapis~~prawdziwy.[/tex]
[tex]b)\\\\\sqrt{1,69\div 169} =0,01\\\\L=\sqrt{1,69\div 169} =\sqrt{\dfrac{169}{100}\cdot \dfrac{1}{169} } =\sqrt{(\dfrac{1}{10} )^{2} } =\dfrac{1}{10}=0,1\\\\P =0,01\\\\L\neq P\\\\Odp:F -zapis~~falszywy.[/tex]
[tex]d)\\\\\sqrt{\dfrac{19}{16} } \div \sqrt{2\dfrac{1}{9} } =\sqrt{\dfrac{16}{9} } \\\\L=\sqrt{\dfrac{19}{16} } \div \sqrt{2\dfrac{1}{9} } =\sqrt{\dfrac{19}{16} } \div \sqrt{\dfrac{19}{9} } =\sqrt{\dfrac{19}{16}\div \dfrac{19}{9}} = \sqrt{\dfrac{19}{16}\cdot \dfrac{9}{19}} =\sqrt{\dfrac{9}{16} } \\\\P=\sqrt{\dfrac{16}{9} }\\\\L\neq P\\\\Odp:Fzapis ~~falszywy.[/tex]
zad.2
[tex]I.\\\\2\sqrt{15} \div \sqrt{5} \cdot \sqrt{3} =2\sqrt{15\div 5\cdot 3} =2\sqrt{15\cdot\dfrac{1}{5} \cdot 3} =2\sqrt{3\cdot 3} =2\sqrt{3^{2} } =2\cdot 3^{2\cdot \frac{1}{2} } =2\cdot 3=6\\\\Odp: B.6[/tex]
[tex]II.\\\\\sqrt{7,5\cdot 10^{5} } \div \sqrt{0,3\cdot 10^{3} }=\dfrac{\sqrt{7,5\cdot 10^{5} }}{\sqrt{0,3\cdot 10^{3} }} =\sqrt{(7,5\div 0,3) \cdot 10^{5-3} } =\sqrt{\dfrac{15}{2} \cdot \dfrac{10}{3} \cdot 10^{2} } =\sqrt{25\cdot 10^{2} } =\sqrt{5^{2} \cdot 10^{2} }=\sqrt{(5\cdot 10)^{2} }=\sqrt{50^{2} } =50^{2\cdot\frac{1}{2} } =50\\\\Odp:~~D.50[/tex]
Rozwiązując zadania korzystałam ze wzorów:
[tex]\sqrt[n]{x^{n} } =x^{n\cdot \frac{1}{n} } =x\\\\\sqrt[n]{x} \cdot \sqrt[n]{y} =\sqrt[n]{x \cdot y} \\\\\dfrac{\sqrt[n]{x} }{\sqrt[n]{y} } =\sqrt[n]{\frac{x}{y} } \\\\x^{n} \cdot y^{n} =(x\cdot y)^{n}[/tex]