Odpowiedź:
[tex]a)\ \ \frac{3}{4}+\frac{2}{3}:(-\frac{8}{12})+(1\frac{1}{2})^2=\frac{3}{4}+\frac{2}{3}:(-\frac{2}{3})+(\frac{3}{2})^2=\frac{3}{4}+\frac{\not2^1}{\not3_{1}}\cdot(-\frac{\not3^1}{\not2_{1}})+\frac{9}{4}=\\\\=\frac{3}{4}-1+\frac{9}{4}=\frac{12}{4}-1=3-1=2[/tex]
[tex]b)\ \ \dfrac{(2-\frac{1}{7})\cdot3}{2+0,6}-8\cdot\frac{3}{7}=\dfrac{(\frac{14}{7}-\frac{1}{7})\cdot3}{2,6}-\frac{24}{7}=\dfrac{\frac{13}{7}\cdot3}{\frac{26}{10}}-\frac{24}{7}=\dfrac{\frac{39}{7}}{\frac{13}{5}}-\frac{24}{7}=\frac{39}{7}:\frac{13}{5}-\frac{24}{7}=\\\\\\=\frac{\not39^3}{7}\cdot\frac{5}{\not13_{1}}-\frac{24}{7}=\frac{15}{7}-\frac{24}{7}=-\frac{9}{7}=-1\frac{2}{7}[/tex]
[tex]c)\ \ (\frac{1}{3}+\frac{2}{7}):(1-\frac{1}{9})+\frac{3}{7}=(\frac{7}{21}+\frac{6}{21}):(\frac{9}{9}-\frac{1}{9})+\frac{3}{7}=\frac{13}{21}:\frac{8}{9}+\frac{3}{7}=\frac{13}{\not21_{7} }\cdot\frac{\not9^3}{8}+\frac{3}{7}=\\\\=\frac{39}{56}+\frac{3}{7}=\frac{39}{56}+\frac{24}{56}=\frac{63}{56}=1\frac{7}{56}=1\frac{1}{8}[/tex]
[tex]d)\ \ (\not2^1\cdot\frac{5}{\not6_{3}})^2-(3^2-\sqrt{1\frac{7}{9}})=(\frac{5}{3})^2-)9-\sqrt{\frac{16}{9}})=\frac{25}{9}-(9-\frac{4}{3})=\frac{25}{9}-(8\frac{3}{3}-1\frac{1}{3})=\\\\=\frac{25}{9}-7\frac{2}{3}=\frac{25}{9}-\frac{23}{3}=\frac{25}{9}-\frac{69}{9}=-\frac{44}{9}=-4\frac{8}{9}[/tex]
[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]
