Zadanie 1 :
[tex]\sqrt{36}=6 \ bo \ 6^2=36\\\\\sqrt{1\frac{7}{9}}=\sqrt{\frac{16}{9}}=\frac{4}{3}=1\frac{1}{3} \ bo \ (1\frac{1}{3})^2=(\frac{4}{3})^2=\frac{16}{9}=1\frac{7}{9}\\\\\sqrt{0,49}=0,7 \ bo \ 0,7^2=0,49\\\\\sqrt[3]{64}=4 \ bo \ 4^3=64\\\\\sqrt{\frac{25}{81}}=\frac{5}{9} \ bo \ (\frac{5}{9})^2=\frac{25}{81}\\\\\sqrt[3]{\frac{8}{125}}=\frac{2}{5} \ bo \ (\frac{2}{5})^3=\frac{8}{125}[/tex]
Zadanie 2 :
[tex](\sqrt{15})^2=15\\\\2\sqrt{64}-\sqrt[3]{27}=2\cdot8-3=16-3=13\\\\4\sqrt{11}+\sqrt{11}=5\sqrt{11}\\\\(\sqrt[3]{22})^3=22\\\\9\sqrt5-4\sqrt5=5\sqrt5[/tex]
