[tex]\displaystyle 8). \frac{1}{4} \cdot \left[ \left(1 \frac{1}{2} \right)^2- \frac{1}{2} + \frac{1}{2} :2 \right]:3= \frac{1}{4} \cdot \left[\left( \frac{2 \cdot 1+1}{2} \right)^2- \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \right]:3= \\ \\ = \frac{1}{4}\cdot \left[ \left( \frac{3}{2} \right)^2- \frac{1}{2} + \frac{1}{4} \right]:3= \frac{1}{4} \cdot \left( \frac{9}{4} - \frac{1}{2} + \frac{1}{4} \right):3= \frac{1}{4}\cdot \frac{9-2+1}{4} \cdot \frac{1}{3} = [/tex]
[tex]\displaystyle =\frac{1}{4} \cdot \frac{8}{4} \cdot \frac{1}{3} = \frac{8}{16} \cdot \frac{1}{3} = \frac{8}{48} ^{(8}= \frac{1}{6} [/tex]
[tex]\displaystyle 9).\left[ \left(-2 \frac{3}{4} \right): \left(- \frac{11}{8}\right) -3 \frac{1}{2} : \frac{7}{4} \right]^{2015}=\\ \\ =\left[\left(- \frac{4 \cdot 2+3}{4} \right) \cdot \left(- \frac{8}{11} \right)- \frac{2 \cdot 3+1}{2} \cdot \frac{4}{7} \right]^{2015}= \\ \\ =\left[\left(- \frac{11}{4} \right) \cdot \left(- \frac{8}{11} \right)- \frac{7}{2} \cdot \frac{4}{7} \right]^{2015}=(2-2)^{2015}=0^{2015}=0[/tex]
[tex]\displaystyle 10).\left(-3 \frac{1}{5} \right):\left(-1 \frac{1}{3} +0,4\right) \cdot 0,1(5)= \\ \\ =\left(- \frac{5 \cdot 3+1}{5} \right):\left(- \frac{3 \cdot 1+1}{3} + \frac{4}{10} \right) \cdot \frac{15-1}{90} = \\ \\ =\left(- \frac{16}{5} \right):\left(- \frac{4}{3} + \frac{4}{10} \right) \cdot \frac{15}{90}=\left(- \frac{16}{5} \right):\left(- \frac{40}{30} + \frac{12}{30}\right) \cdot \frac{14}{90} =[/tex]
[tex]\displaystyle =\left(- \frac{16}{5} \right) : \left(- \frac{28}{30} \right) \cdot \frac{15}{90} =\left(- \frac{16}{5} \right) \cdot \left(- \frac{30}{28} \right) \cdot \frac{14}{90} = \frac{6720}{12600} ^{(840}= \frac{8}{15} [/tex]
[tex]\displaystyle 11).\left[\left( \frac{2}{3} \right)^7 \cdot \left( \frac{2}{3} \right)^{-6}:\left( \frac{2}{3} \right)^2\right]^{-1}+\left[0,(16)-1 \frac{1}{3} \right]- \left( \frac{4}{5} -0,6\right)^{-1}= \\ \\ =\left[\left( \frac{2}{3} \right)^{7-6-2}\right]^{-1}+\left( \frac{16}{99} - \frac{3 \cdot 1+1}{3} \right)-\left( \frac{4}{5} - \frac{6}{10} \right)^{-1}= [/tex]
[tex]\displaystyle =\left[ \left( \frac{2}{3} \right)^{-1}\right]^{-1}+\left( \frac{16}{99} - \frac{4}{3} \right)-\left( \frac{8-6}{10} \right)^{-1}=\\ \\= \left( \frac{3}{2} \right)^{-1}+ \frac{16-132}{99} -\left( \frac{2}{10} \right)^{-1}= \frac{2}{3} - \frac{116}{99} - \frac{10}{2} = \\ \\ = \frac{2}{3} - \frac{116}{99} -5= \frac{66-116-495}{99} =- \frac{545}{99} [/tex]
