Odpowiedź:
zad. 1
[tex]a)6 \sqrt{7} + 2 \sqrt{7} = 8 \sqrt{7} \\ b)7 \sqrt{3} - 5 \sqrt{3} + 2 \sqrt{5} - 4 \sqrt{5} = 2 \sqrt{3} - 2 \sqrt{5} \\ c) - 2 \sqrt{11} + 4 \sqrt{11} + \sqrt{36} = 2 \sqrt{11} + 6[/tex]
[tex]d) \sqrt{49} - \sqrt{64} + \sqrt{16} = 7 - 8 + 4 = 3 \\ e)5 \sqrt{9} + 2 \sqrt{81} = 5 \times 3 + 2 \times 9 = 15 + 18 = 33 \\ f) \sqrt{36} - \sqrt[3]{8} + \sqrt[3]{64} - \sqrt{4} = 6 - 2 + 4 - 2 = 4 + 4 - 2 = 8 - 2 = 6 \\ g)3 \sqrt{2} \times 9 \sqrt{7} = 27 \sqrt{14} \\ h) - 5 \sqrt{2} - 3 \sqrt{2} + 11 \sqrt{2} = - 8 \sqrt{2} + 11 \sqrt{2} = 3 \sqrt{2} \\ i) \sqrt{ {7}^{2} } + \sqrt{ {17}^{2} } = 7 + 17 = 24[/tex]
[tex]j) \sqrt[3]{5} \times \sqrt[3]{25} = \sqrt[3]{5 \times 25} = \sqrt[3]{125} = 5 \\ k) \sqrt{ \frac{36}{100} } = \frac{6}{10} = \frac{3}{5} \\ l) \frac{ \sqrt{9} + \sqrt{121} }{ \sqrt{81} - \sqrt{16} } = \frac{3 + 11}{9 - 4} = \frac{14}{5} [/tex]
zad. 2
[tex]a) \sqrt{8} = \sqrt{4 \times 2} = 2 \sqrt{2} \\ b) \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \\ c) \sqrt{200} = \sqrt{2 \times 100} = 10 \sqrt{2} [/tex]
zad. 3
[tex]a)2 \sqrt{5} = \sqrt{5 \times 4} = \sqrt{20} \\ b)9 \sqrt{2} = \sqrt{81 \times 2} = \sqrt{162} \\ c) \frac{1}{2} \sqrt{8} = \sqrt{8 \times \frac{1}{4} } = \sqrt{2} [/tex]
zad. 4
[tex]a) \frac{1}{ \sqrt{11} } \times \frac{ \sqrt{11} }{ \sqrt{11} } = \frac{1 \sqrt{11} }{11} \\ b) \frac{3}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{3 \sqrt{2} }{2} \\ c) \frac{ \sqrt{3} }{ \sqrt{7} } \times \frac{ \sqrt{7} }{ \sqrt{7} } = \frac{ \sqrt{21} }{7} [/tex]
