[latex]left [sqrt{dfrac{1}{225}} + left(1 dfrac1{14} ight)^{-1} +3 ight]^{ig {0,5}}=left [{dfrac{sqrt1}{sqrt{225}}} + left( dfrac{15}{14} ight)^{-1} +3 ight]^{ig {frac12}}=\\\=left [dfrac{1}{15} + dfrac{14}{15}+3 ight]^{ig {frac12}}=left [dfrac{15}{15} +3 ight]^{ig {frac12}}=left [ 1+3 ight]^{frac12}=sqrt4=2[/latex] [latex]left(4dfrac{17}{27} ight)^{frac23}=left(dfrac{125}{27} ight)^{frac13cdot2}=left[left(dfrac{125}{27} ight)^{frac13} ight]^2=left(sqrt[ig3]{dfrac{125}{27}} ight)^2=left(dfrac{sqrt[3]{125}}{sqrt[3]{27}}} ight)^2=\\\=left(dfrac{5}{3} ight)^2=dfrac{5^2}{3^2}=dfrac{25}{9}=2dfrac79 [/latex] [latex]dfrac{5^{16}-3cdot 5^{15}}{25^5cdot5^6}=dfrac{5^{1+15}-3cdot 5^{15}}{(5^2)^5cdot5^6}=dfrac{5^1cdot5^{15}-3cdot 5^{15}}{5^{2cdot5}cdot5^{5+1}}=dfrac{5^{15}cdot(5-3)}{5^{10}cdot5^5cdot5^1}=\\\=dfrac{5^{15}cdot2}{5^{10+5}cdot5^1}=dfrac{5^{15}cdot2}{5^{15}cdot5}=dfrac25[/latex] [latex] 6^5 : (sqrt3)^{10}=dfrac{6^5}{(sqrt3)^{10}}=dfrac{(2cdot3)^5}{(sqrt3)^{2cdot5}}=dfrac{2^5cdot3^5}{[(sqrt3)^{2}]^5}=dfrac{2^5cdot3^5}{3^5}=2^5=32[/latex] [latex]left(3dfrac18 ight)^{frac12} =sqrt{3dfrac{1}8}=sqrt{dfrac{25}8}=dfrac{sqrt{25}}{sqrt8}=dfrac5{2sqrt2}=dfrac{5cdotsqrt2}{2sqrt2cdotsqrt2}=dfrac{5sqrt2}{4}[/latex]
