Question

помагииите даю 30 баллов решение на листочке

Answer 1

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

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


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