Odpowiedź:
zad12
a)[tex]2\sqrt{11} =\sqrt{2^2*11} =\sqrt{44}[/tex]
b)[tex]\frac{2}{3} *\sqrt{6} =\sqrt{(\frac{2}{3})^2*6 } =\sqrt{\frac{4}{9} *6} =\sqrt{\frac{8}{3} } =\sqrt{2\frac{2}{3} }[/tex]
c)[tex]\frac{4}{5} \sqrt{10} =\sqrt{(\frac{4}{5})^2*10 } =\sqrt{\frac{16}{25}*10 } =\sqrt{\frac{32}{5} } =\sqrt{6\frac{2}{5} }[/tex]
d)[tex]\frac{3}{7} \sqrt{\frac{7}{3} } =\sqrt{(\frac{3}{7})^2*\frac{7}{3} } =\sqrt{\frac{9}{49} *\frac{7}{3} } =\sqrt{\frac{3}{7} }[/tex]
e)[tex]4\sqrt[3]{5} =\sqrt[3]{4^3*5} =\sqrt[3]{64*5} =\sqrt[3]{320}[/tex]
f)[tex]\frac{1}{3} \sqrt[3]{6} =\sqrt[3]{(\frac{1}{3} )^3*6} =\sqrt[3]{\frac{1}{27}*6 } =\sqrt[3]{\frac{2}{9} }[/tex]
Szczegółowe wyjaśnienie:
