Zadanie 1
a)
\frac{(\frac{1}{4})^4*(-8)^4}{5^4*(0,2)^4}=\frac{4^{-4}*8^4}{(5*0,2)^4}=\frac{(2^2)^{-4}*(2^3)^4}{1^4}=\frac{2^{-8}*2^{12}}{1}=2^{-8+12}=2^4=16
b)
\frac{(2^{19}+2^{18})*3^2}{(2^{18}+2^{17})*27}=\frac{2^{17}(2^2+2)*3^2}{2^{17}(2+1)*3^3}=\frac{6*3^2}{3*3^3}=\frac{2*3^*3^2}{3*3^3}=\frac{2*3^3}{3*3^3}=\frac{2}{3}
Zadanie 2
\sqrt{2^{10}}=\sqrt{(2^5)^2}=2^5=32
albo
\sqrt{2^{10}}=(2^{10})^{\frac{1}{2}}=2^5=32
6\sqrt[3]{6}(\sqrt[3]{6})^2 = 6*(\sqrt[3]{6})^{1+2} = 6*\sqrt[3]{6^3}=6*6=36
\frac{\sqrt{49}*7^4}{(\sqrt7)^2*7^2}=\frac{7*7^4}{7*7^2}=\frac{7^4}{7^2}=7^{4-2}=7^2=49
\frac{\sqrt{49}*7^4}{(\sqrt7)^2*7^2}>6\sqrt[3]{6}(\sqrt[3]{6})^2>\sqrt{2^{10}}