Odpowiedź :
[tex]a)\\\sqrt{40} \ \ i \ \ 2\sqrt{7}\\\\2\sqrt{7} =\sqrt{2^{2}\cdot7} = \sqrt{4\cdot7} = \sqrt{28}\\\\\sqrt{40} > \sqrt{28}\\\\\boxed{\sqrt{40} > 2\sqrt{7}}[/tex]
[tex]b)\\3\sqrt{8} \ \ i \ \ \sqrt{63}\\\\3\sqrt{8} = \sqrt{3^{2}\cdot8} = \sqrt{9\cdot8} = \sqrt{72}\\\\\sqrt{72} > \sqrt{63}\\\\\boxed{3\sqrt{8} > \sqrt{63}}[/tex]
[tex]c)\\6\sqrt{5} \ \ i \ \ 2\sqrt{17}\\\\6\cqrt{5} = \sqrt{6^{2}\cdot5} =\sqrt{36\cdot5} = \sqrt{180}\\\\2\sqrt{17} = \sqrt{2^{2}\cdot17} = \sqrt{4\cdot17} = \sqrt{68}\\\\\sqrt{180} > \sqrt{68}\\\\\boxed{6\sqrt{5} > 2\sqrt{17}}[/tex]
[tex]zad.7\\\\a)~~\sqrt{40} ~~i ~~2\sqrt{7} \\\\2\sqrt{7} =\sqrt{2^{2} \cdot 7} =\sqrt{4\cdot 7} =\sqrt{28} \\\\\sqrt{40} >\sqrt{28} \\\\\sqrt{40} >2\sqrt{7}\\\\b)~~3\sqrt{8} ~~i~~\sqrt{63} \\\\3\sqrt{8} =\sqrt{3^{2} \cdot 8} =\sqrt{9\cdot 8} =\sqrt{72} \\\\\sqrt{72} >\sqrt{63} \\\\3\sqrt{8} >\sqrt{63} \\\\c)~~6\sqrt{5} ~~i~~2\sqrt{17} \\\\6\sqrt{5}=\sqrt{6^{2} \cdot 5} =\sqrt{36\cdot 5} =\sqrt{180} \\\\2\sqrt{17} =\sqrt{2^{2} \cdot 17} =\sqrt{4\cdot 17 } =\sqrt{68} \\\\[/tex]
[tex]\sqrt{180} >\sqrt{68} \\\\6\sqrt{5} >2\sqrt{17}[/tex]