Odpowiedź:
[tex]a)\ \ 4,67-(3\frac{5}{6}-\frac{1}{3})=4,67-(3\frac{5}{6}-\frac{2}{6})=4,67-3\frac{3}{6}=4,67-3\frac{1}{2}=4,67-3,5=1,17\\\\b)\ \ 2,8+1\frac{1}{3}\cdot0,9=2,8+\frac{\not4^2}{\not3_{1}}\cdot\frac{\not9^3}{\not10_{5}}=2,8+\frac{6}{5}=2,8+1\frac{1}{5}=2,8+1,2=4\\\\c)\ \ 3,6:\frac{3}{5}-1\frac{1}{5}=3,6:0,6-1\frac{1}{5}=6-1\frac{1}{5}=5\frac{5}{5}-1\frac{4}{5}=4\frac{4}{5}[/tex]
[tex]d)\ \ 0,2^2\cdot(1,6-\frac{2}{5})=(\frac{2}{10})^2\cdot(1,6-0,4)=(\frac{1}{5})^2\cdot1,2=\frac{1}{25}\cdot\frac{12}{10}=\frac{1}{25}\cdot\frac{6}{5}=\frac{6}{125}\\\\e)\ \ 2,8\cdot1\frac{1}{4}-\frac{2}{3}\cdot0,75=\frac{28}{10}\cdot\frac{5}{4}-\frac{2}{3}\cdot\frac{75}{100}=\frac{\not14^7}{\not5_{1}}\cdot\frac{\not5^1}{\not4_{2}}-\frac{\not2^1}{\not3_{1}}\cdot\frac{\not3^1}{\not4_{2}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3[/tex]
[tex]f)\ \ (4,5-1,1)\cdot2\cdot\frac{3}{4}-1=3,4\cdot2\cdot\frac{3}{4}-1=6,8\cdot\frac{3}{4}-1=\frac{68}{10}\cdot\frac{3}{4}-1=\frac{\not34^1^7}{5}\cdot\frac{3}{\not4_{2}}-1=\\\\=\frac{51}{10}-1=5\frac{1}{10}-1=4\frac{1}{10}[/tex]
