[tex]zad.8\\\\a)~~\dfrac{3\sqrt{6} }{2\sqrt{3} } =\dfrac{3 }{2 } \cdot \dfrac{\sqrt{6} }{\sqrt{3} } =\dfrac{3 }{2 } \cdot \sqrt{\dfrac{6}{3} } =\dfrac{3 }{2 }\cdot \sqrt{2}= \dfrac{3\sqrt{2} }{2 }\\\\b)~~\dfrac{\sqrt{12} }{\sqrt{8} }=\sqrt{\dfrac{12}{8} } =\sqrt{\dfrac{3}{2} } =\dfrac{\sqrt{3} }{\sqrt{2} }\cdot \dfrac{\sqrt{2} }{\sqrt{2} }=\dfrac{\sqrt{6} }{2 }\\\\c)~~\dfrac{\sqrt{6} }{3\sqrt{2} }=\dfrac{\sqrt{6} }{3\sqrt{2} }\cdot \dfrac{\sqrt{2} }{\sqrt{2} }=\dfrac{\sqrt{6\cdot 2} }{3\cdot 2 }=[/tex]
[tex]~~~~=\dfrac{\sqrt{12} }{6 }=\dfrac{\sqrt{4\cdot 3} }{6 }=\dfrac{2\sqrt{ 3} }{6 }=\dfrac{\sqrt{3} }{3} \\\\d)~~\dfrac{3\sqrt{2} }{2\sqrt{6} } =\dfrac{3\sqrt{2} }{2\sqrt{6} } \cdot \dfrac{\sqrt{6} }{\sqrt{6} } =\dfrac{3\sqrt{2\cdot 6} }{2\cdot 6 }=\dfrac{3\sqrt{12} }{12} =\dfrac{3\sqrt{4\cdot 3} }{12}=\dfrac{3\cdot 2\sqrt{3} }{12}=\dfrac{6\sqrt{3} }{12}=\dfrac{\sqrt{3} }{2}\\\\\\\\Odp. ~~D[/tex]
[tex]zad.9\\\\\dfrac{1}{\sqrt[3]{2} } +\dfrac{1}{\sqrt[3]{4} } =\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{\sqrt[3]{4}\cdot \sqrt[3]{2}} =\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{\sqrt[3]{4\cdot 2}} =\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{\sqrt[3]{8}} =\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{\sqrt[3]{2^{3} }}=\dfrac{\sqrt[3]{4}+\sqrt[3]{2}}{2}\\\\\\Odp.~~A[/tex]
[tex]zad10\\\\a)\\\\4\sqrt{0,06} =4\sqrt{\dfrac{6}{100} } =4\sqrt{\dfrac{6}{10^{2} } }=\dfrac{4}{10} \sqrt{6} ~~\land ~~\sqrt{6}\approx 2,45\Rightarrow \dfrac{4}{10}\cdot 2,45\approx 0,98\\\\4\sqrt{0,06}<1\\\\b)\\\\0,8\sqrt{2}~~\land ~~ \sqrt{2} \approx1,41 \Rightarrow 0,8\cdot 1,41\approx 1,128\\\\0,8\sqrt{2} >1\\\\c)\\\\2\sqrt{0,4} =2\sqrt{\dfrac{4}{10}} =2\sqrt{\dfrac{1}{5}} =\dfrac{2}{\sqrt{5} } =\dfrac{2\sqrt{5} }{5} =0,4\sqrt{5} ~~\land ~~\sqrt{5} \approx 2,24 \Rightarrow[/tex]
[tex]\Rightarrow ~~0,4\cdot 2,24\approx 0,896\\\\2\sqrt{0,4} <1[/tex]
[tex]d)\\\\0,5\sqrt{10} ~~\land ~~\sqrt{10} \approx 3,16 ~~\Rightarrow ~~0,5\cdot 3,16\approx 1,59\\\\0,5\sqrt{10} >1\\\\Odp. Wyrazenia~~mniejsze~~1~~to~~:~~A~~oraz~~C.[/tex]
[tex]zad.11\\\\\\Liczba~~nieujemna~~to~~taka,~~ktora~~jest~~wieksza~~lub~~rowna~~0.\\\\Wykazac~~nalezy~~5\sqrt{20} -2\sqrt{125} \geq 05\sqrt{20} -2\sqrt{125} =5\sqrt{4\cdot 5} -2\sqrt{25\cdot 5}=5\sqrt{2^{2} \cdot 5} -2\sqrt{5^{2} \cdot 5}=5\cdot 2\cdot \sqrt{5} -2\cdot 5\cdot \sqrt{5} =10\sqrt{5} -10\sqrt{5} =0\\\\5\sqrt{20} -2\sqrt{125} =0~~ \Rightarrow~~5\sqrt{20} -2\sqrt{125} \geq 0~~~~cbdu\\\\[/tex]
