Ktoś pomoże???? Chociaż pare!!!



Ktoś Pomoże Chociaż Pare class=

Odpowiedź :

8.

[tex]\frac{1}{2}\cdot\sqrt[3]{\frac{1}{36}}\cdot\sqrt[3]{36} =\frac{1}{2}\cdot \sqrt[3]{\frac{1}{36}\cdot36} =\frac{1}{2}\cdot \sqrt[3]{1} = \frac{1}{2}\cdot1=\frac{1}{2}\\\\Odp. \ B.[/tex]

9.

[tex]\sqrt{5\frac{1}{3}}\cdot\sqrt{3} = \sqrt{\frac{16}{3}\cdot3} = \sqrt{16} = \sqrt{4^{2}} = 4\\\\\sqrt{9\frac{4}{5}}:\sqrt{5} = \sqrt{\frac{49}{5}:5} = \sqrt{\frac{49}{5}\cdot\frac{1}{5}} = \sqrt{\frac{49}{25}} = \sqrt{(\frac{7}{5})^{2}} = \frac{7}{5} = 1\frac{2}{5}[/tex]

10.

[tex]\sqrt{12}+\sqrt{48} = \sqrt{4\cdot3}+\sqrt{16\cdot3} = \sqrt{4}\cdot\sqrt{3}+\sqrt{16}\cdot\sqrt{3} = 2\sqrt{3}+4\sqrt{3} = 6\sqrt{3}\\\\Odp. \ D.[/tex]

11.

[tex]\sqrt{8}\cdot\sqrt{\frac{2}{3}}\cdot\sqrt{27} =\sqrt{8\cdot\frac{2}{3}\cdot27} = \sqrt{16\cdot9}=\sqrt{144} = \sqrt{12^{2}} = 12[/tex]

[tex]\sqrt{5,4}:\sqrt{2\frac{11}{12}}\cdot\sqrt{7}=\sqrt{\frac{54}{10}:\frac{35}{12}\cdot7} = \sqrt{\frac{27}{5}\cdot\frac{12}{35}\cdot7} = \sqrt{\frac{27}{5}\cdot\frac{12}{5}} = \sqrt{\frac{324}{25}} = \frac{18}{5} = 3\frac{3}{5}[/tex]