Question

Помогите, пожалуйста, решить алгебру.

Answer 1

$a) 5^{log_{frac{1}{5}}frac{1}{2}}+log_{sqrt{2}}frac{4}{sqrt{3}+sqrt{7}}}+log_{frac{1}{2}}frac{1}{10+2sqrt{21}} = \\ 5^{log_{5}2}+log_{sqrt{2}}4-log_{sqrt{2}}(sqrt{3}+sqrt{7})+log_{2}(10+2sqrt{21})=\\ 2+4-log_{sqrt{2}}(sqrt{3}+sqrt{7})+log_{2}(sqrt{3}+sqrt{7})^2 = \\ 6-2log_{2}(sqrt{3}+sqrt{7})+2log_{2}(sqrt{3}+sqrt{7})=6$
          


$b)\ frac{(9^{log_{3}(frac{3sqrt{2}+1}{sqrt{2}})} - 25^{log_{frac{1}{5}frac{sqrt{2}}{3sqrt{2}-1}}})*2^{log_{5}3sqrt{5}}}{3^{log_{5}10}}=\\ frac{(9^{log_{3}(3sqrt{2}+1)-log_{3}sqrt{2}} -25^{-0.5*log_{5}2+log_{5}3sqrt{2}-1}) *2^{log_{5}3+log_{5}sqrt{5}}}{3^{log_{5}10}}=\\ frac{(frac{(3sqrt{2}+1)^2}{2}-5^{-log_{5}2+2log_{5}3sqrt{2}-1})*2^{log_{5}3+0.5log_{5}5} }{3^{log_{5}10}}=\\ frac{12*2^{log_{5}3}}{10^{log_{5}3}}=12*frac{1}{5}^{log_{5}3}=12*5^{-log_{5}3}=12*frac{1}{3}=4$ 



$(sqrt[3]{3})^{frac{-1}{log_{64}3^{-1}}}+0.25*log_{5}(12-2sqrt{35})-log_{frac{1}{25}}frac{25}{sqrt{7}-sqrt{5}} \\ 4 +0.5*log_{5}(sqrt{7}-sqrt{5})-log_{frac{1}{25}}25+log_{frac{1}{25}}(sqrt{7}-sqrt{5})=\\ 4+1=5$





$frac{7^{log_{3}15}}{5^{log_{3}7+log_{5}2}}}+3^{log_{sqrt{3}}(2-frac{1}{sqrt{2}})}+4^{log_{frac{1}{2}}frac{sqrt{2}}{2sqrt{2}+1}} = \\ frac{7^{log_{3}15}}{5^{log_{3}7}*2}+(2-frac{1}{sqrt{2}})^2+2^{log_{0.5}sqrt{2}-log{0.5}(2sqrt{2}+1)}=\\ frac{7}{2} + (frac{9}{2}-2sqrt{2}) + frac{frac{1}{sqrt{2}}}{(1+2sqrt{2})^{-1}}}=\\ frac{16-4sqrt{2}}{2}+frac{frac{1}{sqrt{2}}}{frac{1}{(1+2sqrt{2})}}=\\ frac{16-4sqrt{2}}{2} + frac{sqrt{2}+4}{2}=frac{20-3sqrt{2}}{2}$