Odpowiedź :
zad.4
[tex]\sqrt{36} = \sqrt{6^{2}} = 6\\\\\sqrt{16} = \sqrt{4^{2}} = 4\\\\\sqrt{25} = \sqrt{5^{2}} = 5\\\\\sqrt[3]{64} = \sqrt[3]{4^{3}} = 4\\\\\sqrt[3]{0,001} = \sqrt[3]{0,1^{3}} = 0,1[/tex]
zad.4
[tex]\sqrt{36} = \sqrt{6^{2}} = 6\\\\\sqrt{16} = \sqrt{4^{2}} = 4\\\\\sqrt{25} = \sqrt{5^{2}} = 5\\\\\sqrt[3]{64} = \sqrt[3]{4^{3}} = 4\\\\\sqrt[3]{0,001} = \sqrt[3]{0,1^{3}} = 0,1[/tex]