[tex]a)\ \ 5^3\cdot(\frac{2}{5})^3-(\frac{1}{3})^3\cdot9^2:3=(\not5^1\cdot\frac{2}{\not5_{1} })^3-(\frac{1}{3})^3\cdot(3^2)^2:3^1=2^3-(\frac{1}{3})^3\cdot3^4:3^1=\\\\=8-(\frac{1}{3})^3\cdot3^{4-1}=8-(\frac{1}{3})^3\cdot3^3=8-(\frac{1}{\not3_{1}}\cdot\not3^1)^3=8-1^3=8-1=7[/tex]
[tex]b)\ \ (-1^5)^2:(0,5)^2-[-\frac{4^2}{3}+(1\frac{1}{3})^2]^1=(-1)^2:(0,5)^2-[-\frac{16}{3}+(\frac{4}{3})^2]=\\\\=1^2:0,5^2-(-\frac{16}{3}+\frac{16}{9})=(1:0,5)^2-(-\frac{48}{9}+\frac{16}{9})=2^2-(-\frac{32}{9})=\\\\=4+\frac{32}{9}=4+3\frac{5}{9}=7\frac{5}{9}[/tex]
[tex]c)\ \ \dfrac{(\frac{1}{9})^3\cdot27^3}{(-\frac{1}{2})^6}:\dfrac{(-\frac{2}{3})^2}{(\frac{1}{2})^4}=\dfrac{(\frac{1}{\not9_{1}}\cdot\not27^3)^3}{(\frac{1}{2})^6}:\dfrac{(\frac{2}{3})^2}{(\frac{1}{2})^4}=\dfrac{3^3}{(\frac{1}{2})^6}\ \cdot\dfrac{(\frac{1}{2})^4}{(\frac{2}{3})^2}=\dfrac{3^3}{(\frac{1}{2})^{6-4}}\ \cdot\dfrac{1}{\frac{4}{9}}=[/tex]
[tex]=\dfrac{27}{(\frac{1}{2})^2}\cdot1:\frac{4}{9}=\dfrac{27}{\frac{1}{4}}\cdot1\cdot\frac{9}{4}=27:\frac{1}{4}\cdot\frac{9}{4}=27\cdot\not4\cdot\frac{9}{\not4}=27\cdot9=243[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
[tex]a)\ \ 5^3\cdot(\frac{2}{5})^3-(\frac{1}{3})^3\cdot9^2:3=(\not5^1\cdot\frac{2}{\not5_{1} })^3-(\frac{1}{3})^3\cdot(3^2)^2:3^1=2^3-(\frac{1}{3})^3\cdot3^4:3^1=\\\\=8-(\frac{1}{3})^3\cdot3^{4-1}=8-(\frac{1}{3})^3\cdot3^3=8-(\frac{1}{\not3_{1}}\cdot\not3^1)^3=8-1^3=8-1=7[/tex]
[tex]b)\ \ (-1^5)^2:(0,5)^2-[-\frac{4^2}{3}+(1\frac{1}{3})^2]^1=(-1)^2:(0,5)^2-[-\frac{16}{3}+(\frac{4}{3})^2]=\\\\=1^2:0,5^2-(-\frac{16}{3}+\frac{16}{9})=(1:0,5)^2-(-\frac{48}{9}+\frac{16}{9})=2^2-(-\frac{32}{9})=\\\\=4+\frac{32}{9}=4+3\frac{5}{9}=7\frac{5}{9}[/tex]
[tex]c)\ \ \dfrac{(\frac{1}{9})^3\cdot27^3}{(-\frac{1}{2})^6}:\dfrac{(-\frac{2}{3})^2}{(\frac{1}{2})^4}=\dfrac{(\frac{1}{\not9_{1}}\cdot\not27^3)^3}{(\frac{1}{2})^6}:\dfrac{(\frac{2}{3})^2}{(\frac{1}{2})^4}=\dfrac{3^3}{(\frac{1}{2})^6}\ \cdot\dfrac{(\frac{1}{2})^4}{(\frac{2}{3})^2}=\dfrac{3^3}{(\frac{1}{2})^{6-4}}\ \cdot\dfrac{1}{\frac{4}{9}}=[/tex]
[tex]=\dfrac{27}{(\frac{1}{2})^2}\cdot1:\frac{4}{9}=\dfrac{27}{\frac{1}{4}}\cdot1\cdot\frac{9}{4}=27:\frac{1}{4}\cdot\frac{9}{4}=27\cdot\not4\cdot\frac{9}{\not4}=27\cdot9=243[/tex]