[tex]e)\ \ [(-\frac{2}{3})^2:\frac{8}{9}]^2-(\frac{3}{4}-1\frac{1}{8})=(\frac{\not4^1}{\not9_{1} }\cdot\frac{\not9^1}{\not8_{2}})^2-(\frac{3}{4}-\frac{9}{8})=(\frac{1}{2})^2-(\frac{6}{8}-\frac{9}{8})=\frac{1}{4}-(-\frac{3}{8})=\\\\=\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}[/tex]
[tex]f)\ \ 1,6\cdot\frac{7}{16}-\frac{2,7}{3^3}-\frac{5\cdot0,7}{0,35}=\frac{\not16^1}{10}\cdot\frac{7}{\not16_{1}}-\frac{2,7}{27}-\frac{3,5}{0,35}=\frac{7}{10}-0,1-10=0,7-0,1-10=\\\\=0,6-10=-9,4[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
[tex]e)\ \ [(-\frac{2}{3})^2:\frac{8}{9}]^2-(\frac{3}{4}-1\frac{1}{8})=(\frac{\not4^1}{\not9_{1} }\cdot\frac{\not9^1}{\not8_{2}})^2-(\frac{3}{4}-\frac{9}{8})=(\frac{1}{2})^2-(\frac{6}{8}-\frac{9}{8})=\frac{1}{4}-(-\frac{3}{8})=\\\\=\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}[/tex]
[tex]f)\ \ 1,6\cdot\frac{7}{16}-\frac{2,7}{3^3}-\frac{5\cdot0,7}{0,35}=\frac{\not16^1}{10}\cdot\frac{7}{\not16_{1}}-\frac{2,7}{27}-\frac{3,5}{0,35}=\frac{7}{10}-0,1-10=0,7-0,1-10=\\\\=0,6-10=-9,4[/tex]