Question

помагииите даю 30 баллов решение на листочке

Answer 1

$displaystyle 11.\a)..2*4^{-2}-(frac{2}{3})^{-3}-(frac{1}{5})^{0}+16*8^{-2}+(frac{2}{5})^{-3}+((frac{7}{9})^{-1})^{0}=\\=frac{2}{16}-(frac{3}{2})^{3}-1+frac{16}{64}+(frac{5}{2})^{3}+1=frac{1}{8}-frac{27}{8}+frac{2}{8}+frac{125}{8}=frac{101}{8}=\\=12frac{5}{8}=12,625;$

$displaystyle b)..12*3^{-3}-(frac{3}{5})^{-2}-(frac{1}{2})^{0}-45*9^{-2}+((frac{4}{7})^{-1})^{0}+4*(frac{2}{5})^{-2}=\\=frac{12}{27}-(frac{5}{3})^{2}-1-frac{45}{81}+1+4*(frac{5}{2})^{2}= frac{12}{27}- frac{75}{27}- frac{15}{27}+25= \ \ 25- frac{78}{27}=25-2 frac{8}{9}=22 frac{1}{9};$


$displaystyle 12. \ a)..( frac{3}{5})^{-8}*( frac{5}{3})^{-6}-3^{-11}*( frac{1}{3})^{-9}=( frac{5}{3})^{8}*( frac{5}{3})^{-6}-( frac{1}{3})^{11}*( frac{1}{3})^{-9}= \ \=( frac{5}{3})^{2}-( frac{1}{3})^{2}= frac{24}{9}=2 frac{2}{3}; \ \ \ b)..2^{-12}*( frac{1}{2})^{-10}+( frac{2}{3})^{-5}*( frac{3}{2})^{-3}=2^{-12}*2^{10}+( frac{3}{2})^{5}*( frac{3}{2})^{-3}= \ \ =2^{-2}+( frac{3}{2})^{2}= frac{1}{4}+ frac{9}{4}= frac{10}{4}=2,5;$