Odpowiedź:
[tex]\sqrt{40} = \sqrt{2^{2}*10}= \sqrt{2^{2} } \sqrt{10} = 2\sqrt{10}\\\\\sqrt{810} = \sqrt{9^{2}*10}= \sqrt{9^{2} } \sqrt{10} = 9\sqrt{10}\\\\\sqrt{\frac{5}{49} } = \frac{\sqrt{} 5}{7} \\\sqrt[3]{243} = 3\sqrt[3]{3^{2} } = 3\sqrt[3]{9} \ \\\\\sqrt[3]{-128} = -\sqrt[3]{128} = -2^{2} \sqrt[3]{2} = -4\sqrt[3]{2} \\\\\sqrt[3]{\frac{11}{125} } = \frac{\sqrt[3]{} 11}{5}[/tex]
