Question

MATeMAtyka 1 podst. i roz.:
Zadanie 5 Strona 17: Oblicz.
(ja potrzebuję tylko: e,f,g,h)

Answer 1

[tex]e) \frac{3}{5} -(1,4*\frac{5}{14} -\frac{0,9}{3^{2} } ):(-\frac{1}{5}) =\\\frac{3}{5} -(\frac{14}{10} *\frac{5}{14} -\frac{\frac{9}{10} }{9 } ):(-\frac{1}{5})=\\\frac{3}{5} -(\frac{1}{2}-\frac{9}{10}*\frac{1}{9} )*(-5)=\\\frac{3}{5}-(\frac{1}{2}-\frac{1}{10} )*(-5)=\\\frac{3}{5}-(\frac{5}{10}-\frac{1}{10} )*(-5)=\\\frac{3}{5}-\frac{4}{10}*(-5)=\\\frac{3}{5}-(-\frac{4}{2})=\\\frac{3}{5}+\frac{4}{2}=\frac{3}{5}+2=2\frac{3}{5}[/tex]

[tex]f) \frac{5\frac{3}{4}-8\frac{1}{3} }{\frac{1}{2}-(\frac{2}{3})^2} -\frac{2}{3}:(-\frac{4}{27})=\\\frac{5\frac{3}{4}-8\frac{1}{3} }{\frac{1}{2}-\frac{4}{9}} -\frac{2}{3}*(-\frac{27}{4})=\\\frac{5\frac{3}{4}-8\frac{1}{3} }{\frac{1}{2}-\frac{4}{9}} -(-\frac{9}{2})=\\\frac{5\frac{9}{12}-8\frac{4}{12} }{\frac{9}{18}-\frac{8}{18}} +\frac{9}{2}=\\\frac{-3\frac{5}{12} }{\frac{1}{18} } +\frac{9}{2}=\\{-3\frac{5}{12} }*\frac{18}{1} +\frac{9}{2}=\\-3\frac{5*18}{12*1} +\frac{9}{2}=\\[/tex]

[tex]-3\frac{5*3}{2*1}+\frac{9}{2}=\\-3\frac{15}{2}+\frac{9}{2}=\\-10\frac{1}{2}+4\frac{1}{2}=-6[/tex]

[tex]g)\frac{4}{5} :(\frac{1}{2}-\frac{1}{5})-(\frac{2}{3}-(\frac{3}{8}-1\frac{1}{3}))=\\\frac{4}{5} :(\frac{5}{10}-\frac{2}{10} )-(\frac{2}{3}-(\frac{3}{8}-\frac{4}{3}) )=\\\frac{4}{5} :\frac{3}{10} -(\frac{2}{3} -\frac{9-32}{24} )=\\\frac{4}{5} *\frac{10}{3} -(\frac{16}{24} -\frac{-23}{24} )=\\\frac{8}{3} -(\frac{16}{24} +\frac{23}{24} )=\\\frac{64}{24} -\frac{39}{24} =\frac{25}{24} =1\frac{1}{24}[/tex]

[tex]h) ((-1\frac{1}{2}) ^2-(-1\frac{1}{3})^2):4\frac{1}{4} = \\ ((-\frac{3}{2}) ^2-(-\frac{4}{3})^2):\frac{17}{4}=\\ (-\frac{9}{4}-(-\frac{16}{9})):\frac{17}{4}=\\ (-\frac{9}{4}+\frac{16}{9}):\frac{17}{4}=\\(-\frac{81}{36}+\frac{64}{36}):\frac{17}{4}=\\-\frac{17}{36}*\frac{4}{17}=-\frac{1}{9}[/tex]