Odpowiedź:
[tex]1.4\\\\3\frac{3}{4}\cdot1,2+(-1\frac{1}{2}):2\frac{2}{5}-2^{-2}=\frac{\not15^3}{\not4_{1} }\cdot\frac{\not12^3}{\not10_{2}}-1\frac{1}{2}:\frac{12}{5}-(\frac{1}{2})^2=\frac{9}{2}-\frac{\not3^1}{2}\cdot\frac{5}{\not12_{4}}-\frac{1}{4}=\\\\=\frac{9}{2}-\frac{5}{8}-\frac{1}{4}=\frac{36}{8}-\frac{5}{8}-\frac{2}{8}=\frac{29}{8}=3\frac{5}{8}[/tex]
[tex]1.5\\\\8:\left(0,3-\dfrac{2-\frac{1}{\not10_{5}}\cdot\not4^2}{4}\right)=8:\left(0,3-\dfrac{2-\frac{2}{5}}{4}\right)=8:\left(0,3-\dfrac{\frac{10}{5}-\frac{2}{5}}{4}\right)=\\\\\\=8:\left(0,3-\dfrac{\frac{8}{5}}{4}\right)=8:(0,3-\frac{8}{5}:4)=8:(0,3-\frac{\not8^2}{5}\cdot\frac{1}{\not4_{1}})=8:(\frac{3}{10}-\frac{2}{5})=\\\\\\=8:(\frac{3}{10}-\frac{4}{10})=8:(-\frac{1}{10})=8\cdot(-10)=-80[/tex]
