Przed wykonywaniem dodawania i odejmowania należy sprowadzić liczby do wspólnego mianownika.

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a)

$5\frac{2^{\cdot 7}}{3_{\cdot 7}}+3\frac{4^{\cdot 3}}{7_{\cdot 3}}=5\frac{14}{21}+3\frac{12}{21}=8\frac{26}{21}=9\frac{5}{21}$   

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$1\frac{4^{\cdot 6}}{5_{\cdot 6}}+5\frac{5^{\cdot 5}}{6_{\cdot 5}}=1\frac{24}{30}+5\frac{25}{30}=6\frac{49}{30}=7\frac{19}{30}$    

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$4\frac{7^{\cdot 2}}{9_{\cdot 2}}+\frac{1^{\cdot 3}}{6_{\cdot 3}}=4\frac{14}{18}+\frac{3}{18}=4\frac{17}{18}$  

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$3\frac{7^{\cdot 3}}{{10}_{\cdot 3}}+1\frac{4^{\cdot 2}}{{15}_{\cdot 2}}=3\frac{21}{30}+1\frac{8}{30}=4\frac{29}{30}$  

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b)

$3\frac{5^{\cdot 1}}{6_{\cdot 1}}-1\frac{2^{\cdot 2}}{3_{\cdot 2}}=3\frac{5}{6}-1\frac{4}{6}=2\frac{1}{6}$  

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$6\frac{3^{\cdot 10}}{7_{\cdot 10}}-6\frac{1^{\cdot 7}}{{10}_{\cdot 7}}=6\frac{30}{70}-6\frac{7}{70}=\frac{23}{70}$  

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$4\frac{5^{\cdot 2}}{{12}_{\cdot 2}}-\frac{3^{\cdot 3}}{8_{\cdot 3}}=4\frac{10}{24}-\frac{9}{24}=4\frac{1}{24}$  

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$\frac{7^{\cdot 2}}{9_{\cdot 2}}-\frac{1^{\cdot 3}}{6_{\cdot 3}}=\frac{14}{18}-\frac{3}{18}=\frac{11}{18}$  

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$1\frac{3^{\cdot 4}}{5_{\cdot 4}}-1\frac{1^{\cdot 5}}{4_{\cdot 5}}=1\frac{12}{20}-1\frac{5}{20}=\frac{7}{20}$   

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c)

$7\frac{2^{\cdot 7}}{5_{\cdot 7}}-2\frac{5^{\cdot 5}}{7_{\cdot 5}}=7\frac{14}{35}-2\frac{25}{35}=6\frac{49}{35}-2\frac{25}{35}=4\frac{24}{35}$   

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$3-1\frac{7}{9}=2\frac{9}{9}-1\frac{7}{9}=1\frac{2}{9}$  

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$1\frac{3^{\cdot 2}}{{11}_{\cdot 2}}-\frac{1^{\cdot 11}}{2_{\cdot 11}}=1\frac{6}{22}-\frac{11}{22}=\frac{28}{22}-\frac{11}{22}=\frac{17}{22}$  

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$5\frac{3^{\cdot 4}}{{10}_{\cdot 4}}-2\frac{7^{\cdot 5}}{8_{\cdot 5}}=5\frac{12}{40}-2\frac{35}{40}=4\frac{52}{40}-2\frac{35}{40}=2\frac{17}{40}$        

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$1\frac{3^{\cdot 3}}{4_{\cdot 3}}-\frac{11}{12}=1\frac{9}{12}-\frac{11}{12}=\frac{21}{12}-\frac{11}{12}=\frac{{10}^{:2}}{{12}_{:2}}=\frac{5}{6}$   

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d)

$8-2\frac{7}{12}=7\frac{12}{12}-2\frac{7}{12}=5\frac{5}{12}$    

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$6\frac{5^{\cdot 8}}{9_{\cdot 8}}-5\frac{3^{\cdot 9}}{8_{\cdot 9}}=6\frac{40}{72}-5\frac{27}{72}=1\frac{13}{72}$  

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$4\frac{5^{\cdot 2}}{6_{\cdot 2}}+8\frac{3^{\cdot 3}}{4_{\cdot 3}}=4\frac{10}{12}+8\frac{9}{12}=12\frac{19}{12}=13\frac{7}{12}$   

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$2\frac{7^{\cdot 2}}{{12}_{\cdot 2}}+1\frac{5^{\cdot 3}}{8_{\cdot 3}}=2\frac{14}{24}+1\frac{15}{24}=3\frac{29}{24}=4\frac{5}{24}$  

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$1\frac{9^{\cdot 3}}{{16}_{\cdot 3}}-1\frac{1^{\cdot 8}}{6_{\cdot 8}}=1\frac{27}{48}-1\frac{8}{48}=\frac{19}{48}$