[tex]zad.i\\\\3\cdot \sqrt[3]{8} + \frac{1}{2} =3\cdot \sqrt[3]{2^{3} } + \frac{1}{2}= 3\cdot (2^{3} )^{\frac{1}{3} } + \frac{1}{2}= 3\cdot 2^{3\cdot\frac{1}{3} } +\frac{1}{2}= 3\cdot 2+\frac{1}{2}= 6 + \frac{1}{2}= 6\frac{1}{2}[/tex]
[tex]zad.j\\\\\frac{3}{4} \cdot \frac{\sqrt{25} }{7} =\frac{3}{4} \cdot \frac{\sqrt{5^{2} } }{7} =\frac{3}{4} \cdot \frac{(5^{2} )^{\frac{1}{2} } }{7} = \frac{3}{4} \cdot \frac{5 }{7} =\frac{15}{28}[/tex]
[tex]zad.k\\\\(-\frac{2}{7} )\div 1\frac{1}{3} = (-\frac{2}{7}) \div \frac{4}{3} = (-\frac{2}{7} )\cdot \frac{3}{4} = - \frac{6}{28} = - \frac{3}{14}[/tex]
korzystałam ze wzorów:
[tex]\sqrt[n]{x} = x^{\frac{1}{n} } \\\\(x^{n} )^{m} = x^{n \cdot m}[/tex]
Pamiętamy również:
[tex](-) \cdot(+) = (-)\\\\a \div b = a\cdot \frac{1}{b} ,~~ zal. b\neq 0[/tex]
