Cześć ;-)
1)
[tex]1\frac{3}{4}\cdot1\frac{1}{7}-3:4\frac{1}{2}=\frac{\not7^1}{\not4_1}\cdot\frac{\not8^2}{\not7_1}-3:\frac{9}{2}=\\\\=2-\not3^1\cdot\frac{2}{\not9_3}=2-\frac{2}{3}=1\frac{1}{3}[/tex]
2)
[tex]1\frac{5}{7}\cdot(2\frac{3}{4}-1\frac{2}{3}:4)=\frac{12}{7}\cdot(2\frac{3}{4}-\frac{5}{3}\cdot\frac{1}{4})=\\\\=\frac{12}{7}\cdot(2\frac{3}{4}-\frac{5}{12})=\frac{12}{7}\cdot(2\frac{9}{12}-\frac{5}{12})=\\\\=\frac{12}{7}\cdot2\frac{4}{12}=\frac{\not12^1}{\not7_1}\cdot\frac{\not28^4}{\not12_1}=4[/tex]
3)
[tex]3\frac{1}{3}\cdot(2\frac{1}{4}-1\frac{1}{2}:5)=\frac{10}{3}\cdot(2\frac{1}{4}-\frac{3}{2}\cdot\frac{1}{5})=\\\\=\frac{10}{3}\cdot(2\frac{1}{4}-\frac{3}{10})=\frac{10}{3}\cdot(2\frac{5}{20}-\frac{6}{20})=\\\\=\frac{10}{3}\cdot(1\frac{25}{20}-\frac{6}{20})=\frac{10}{3}\cdot1\frac{19}{20}=\frac{\not10^1}{\not3_1}\cdot\frac{\not39^{13}}{\not20_2}=\frac{13}{2}=6\frac{1}{2}[/tex]
4)
[tex]\dfrac{3\frac{1}{4}+\frac{1}{2}}{3\frac{1}{3}\cdot(1\frac{5}{6}-1\frac{1}{12})}=\dfrac{3\frac{1}{4}+\frac{2}{4}}{\frac{10}{3}\cdot(1\frac{10}{12}-1\frac{1}{12})}=\dfrac{3\frac{3}{4}}{\frac{\not10^5}{\not3_1}\cdot\frac{\not9^3}{\not12_6}}=\\\\\\=\dfrac{\frac{15}{4}}{\frac{15}{6}}=\frac{\not15^1}{\not4_2}\cdot\frac{\not6^3}{\not15_1}=\frac{3}{2}=1\frac{1}{2}[/tex]
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