14.
[tex]a) \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}[/tex]
[tex]b) \sqrt{2^{5}} = \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}[/tex]
[tex]c) \sqrt[3]{56} = \sqrt[3]{8 \cdot 7} = 2\sqrt[3]{7}[/tex]
[tex]d) \sqrt[3]{3^{5}} = \sqrt[3]{243} = \sqrt[3]{27 \cdot 9} = 3\sqrt[3]{9}[/tex]
15.
[tex]0.3\sqrt{5} = \sqrt{0.09\cdot 5} = \sqrt{0.45}[/tex]
prawda
19.
[tex]a)\sqrt[3]{1000} + \sqrt{16} = 10 + 4 = 14[/tex]
[tex]b) \sqrt[3]{5^{2}+10^{2}} = \sqrt[3]{25 + 100} = \sqrt[3]{125} = 5[/tex]
[tex]c) \sqrt{5^{2}} + \sqrt[3]{17^{3}} = 5 + 17 = 22[/tex]
[tex]d) \frac{\sqrt{72}+\sqrt{2} }{\sqrt{2} } = \frac{\sqrt{72:2}+\sqrt{2:2} }{1 }= \sqrt{36} + \sqrt{1} = 6 + 1 = 7[/tex]
[tex]e) \sqrt[3]{2} \cdot \sqrt[3]{32} + \sqrt{2} \cdot \sqrt{8} = \sqrt[3]{2 \cdot 32} + \sqrt{2 \cdot 8} = \sqrt[3]{64} + \sqrt{16} = 4+4 = 8[/tex]
20.
[tex]a)\frac{9}{\sqrt{3} } \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}[/tex]
[tex]b) \frac{-3\sqrt{10} }{2\sqrt{5} } \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{-3\sqrt{10\cdot 5} }{2 \cdot 5} = \frac{-3\sqrt{50} }{10} = \frac{-3\sqrt{25 \cdot 2}}{10 } = \frac{-15\sqrt{2} }{10} = \frac{-3\sqrt{2} }{2}[/tex]
21.
[tex]4\sqrt{3} < 8[/tex] więc 8 jest długością przeciwprostokątnej w tym trójkącie
Pole trójkąta: [tex]\frac{a \cdot h}{2}[/tex]
a = 4
h = 4√3
[tex]P = \frac{4\cdot 4\sqrt{3} }{2} = \frac{16\sqrt{3} }{2} = 8\sqrt{3}[/tex]
22.
[tex]a)\sqrt{8} + 4\sqrt{2} = \sqrt{4 \cdot 2} + 4\sqrt{2} = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}[/tex]
[tex]b) \frac{\sqrt{54}-\sqrt{24} }{\sqrt{6} } = \frac{\sqrt{9 \cdot 6} - \sqrt{4 \cdot 6} }{\sqrt{6} } = \frac{3\sqrt{6} - 2\sqrt{6} }{\sqrt{6} } = \frac{1\sqrt{6} }{\sqrt{6} } = 1[/tex]
[tex]c) \sqrt[3]{128} - \sqrt[3]{-16} = \sqrt[3]{64 \cdot 2} - \sqrt[3]{-8 \cdot 2} = 4\sqrt[3]{2} - (-2\sqrt[3]{2} ) = 4\sqrt[3]{2} + 2\sqrt[3]{2}= 6\sqrt[3]{2}[/tex]
23.
[tex]\frac{2\sqrt{18}-\sqrt{32} }{2\sqrt{2} } = \frac{2\sqrt{9\cdot 2} - \sqrt{16 \cdot 2} }{2\sqrt{2} } = \frac{6\sqrt{2} - 4\sqrt{2} }{2\sqrt{2} } = \frac{2\sqrt{2}}{2\sqrt{2}} = 1[/tex]
24.
[tex](11x^{3}y^{2})^{2}:(121x^{3}y^{6}) = \frac{121 \cdot x^{6} \cdot y^{4}}{121\cdot x^{3}\cdot y^{6}} = \frac{1 \cdot x^{3}\cdot 1}{1 \cdot 1 \cdot y^{2}} = \frac{x^{3}}{y^{2}}[/tex]
[tex]\frac{3^{3}}{(\sqrt{2})^{2} } = \frac{27}{2} = 13.5[/tex]
