Question

simplifique os radicais encontre o perímetro: ajudem pfvr​

Answer 1

$\sqrt{48} = 4 \sqrt{3} \\ \\ \sqrt{40} = 2 \sqrt{10} \\ \\ \sqrt{243 } = 9 \sqrt{3} \\ \\ \sqrt[3]{72} = 2 \sqrt[3]{9} \\ \\ \sqrt{98} = 7 \sqrt{2} \\ \\ \sqrt{363} = 11 \sqrt{3} \\ \\ \sqrt[3]{108} = 3 \sqrt[3]{4} \\ \\ \sqrt[5]{224} =2 \sqrt[5]{7}$

Att: Cr7