[tex]a)\ \ \sqrt{2}=2^{\frac{1}{2}}\\\\\\b)\ \ \sqrt[3]{2}=2^{\frac{1}{3}}\\\\\\c)\ \ \sqrt{0,5}=0,5^{\frac{1}{2}}=(\frac{1}{2})^{\frac{1}{2}}=(2^{-1})^{\frac{1}{2}}=2^{-\frac{1}{2}}\\\\\\d)\ \ \sqrt[3]{4}=4^{\frac{1}{3}}=(2^2)^{\frac{1}{3}}=2^{2\cdot\frac{1}{3}}=2^{\frac{2}{3}}\\\\\\e)\ \ \sqrt{512}=\sqrt{256\cdot2}=\sqrt{256}\cdot\sqrt{2}=16\sqrt{2}=2^4\cdot2^{\frac{1}{2}}=2^{4+\frac{1}{2}}=2^{\frac{9}{2}}[/tex]
[tex]f)\ \ \sqrt[3]{1024}=\sqrt[3]{512\cdot2}=\sqrt[3]{512}\cdot\sqrt[3]{2}=8\sqrt[3]{2}=2^3\cdot2^{\frac{1}{3}}=2^{3+\frac{1}{3}}=2^{\frac{10}{3}}\\\\\\g)\ \ 2\sqrt{2}=2\cdot2^{\frac{1}{2}}=2^1\cdot2^{\frac{1}{2}}=2^{1+\frac{1}{2}}=2^{\frac{3}{2}}\\\\\\h)\ \ 2\sqrt[3]{2}=2\cdot2^{\frac{1}{3}}=2^1\cdot2^{\frac{1}{3}}=2^{1+\frac{1}{3}}=2^{\frac{4}{3}}[/tex]
Zastosowano wzory
[tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}\\\\\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\ \ \ \ dla\ \ a,b\geq 0\\\\a^m\cdot a^n=a^{m+n}[/tex]
