[tex]a) \ \sqrt{1\dfrac{9}{16}}\cdot(-2)^{2}+1,2\cdot\dfrac{5}{12}=\sqrt{\dfrac{16\cdot1+9}{16}}\cdot4+\dfrac{12}{10}\cdot\dfrac{5}{12}=\\\\=\sqrt{\dfrac{25}{16}}\cdot4+\dfrac{\not\!\!12\ \cdot\not5}{\not\!\!10_{2}\ \cdot\not\!\!12}=\sqrt{\dfrac{5^{2}}{4^{2}}}\cdot4+\dfrac{1}{2}=\dfrac{5}{\not\!4}\ \cdot\not\!4+\dfrac{1}{2}=5+\dfrac{1}{2}=\boxed{5\dfrac{1}{2}}[/tex]
[tex]b) \ 2\dfrac{1}{2}-\dfrac{1}{2}\cdot\sqrt[3]{3\dfrac{3}{8}}+\dfrac{1}{2}=2\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2}\cdot\sqrt[3]{\dfrac{27}{8}}=3-\dfrac{1}{2}\cdot\sqrt[3]{\dfrac{3^{3}}{2^{3}}}=\\\\=3-\dfrac{1}{2}\cdot\dfrac{3}{2}=3-\dfrac{1\cdot3}{2\cdot2}=3-\dfrac{3}{4}=2\dfrac{4}{4}-\dfrac{3}{4}=\boxed{2\dfrac{1}{4}}[/tex]
[tex]c) \ \dfrac{4}{3}-(-1\dfrac{1}{3})^{2}:0,9=\dfrac{4}{3}-(-\dfrac{4}{3})^{2}:\dfrac{9}{10}=\dfrac{4}{3}-\dfrac{16}{9}\cdot\dfrac{10}{9}=\\\\=\dfrac{4}{3}-\dfrac{160}{81}=\dfrac{4\cdot27}{3\cdot27}-\dfrac{160}{81}=\dfrac{108}{81}-\dfrac{160}{81}=\boxed{-\dfrac{52}{81}}[/tex]
[tex]d) \ 0,8\cdot\dfrac{3}{2^{3}}-\dfrac{(-2)^{3}}{4}\cdot0,1=\dfrac{\not\!8}{10}\cdot\dfrac{3}{\not\!8}-\dfrac{-\!\!\!\not8^{2}}{\not\!4}\cdot\dfrac{1}{10}=\dfrac{3}{10}-(-2)\cdot\dfrac{1}{10}=\\\\=\dfrac{3}{10}-(-\dfrac{2}{10})=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{5}{10}=\boxed{\dfrac{1}{2}}[/tex]
[tex]e) \ (1\dfrac{1}{2})^{2}\cdot(-3)^{2}-1,9^{0}\cdot\dfrac{1}{3}=(\dfrac{3}{2})^{2}\cdot9-1\cdot\dfrac{1}{3}=\dfrac{9}{4}\cdot9-\dfrac{1}{3}=\\\\=\dfrac{81}{4}-\dfrac{1}{3}=20\dfrac{1}{4}-\dfrac{1}{3}=20\dfrac{3}{12}-\dfrac{4}{12}=19\dfrac{15}{12}-\dfrac{4}{12}=\boxed{19\dfrac{11}{12}}[/tex]
