matematyka zadanie trzecie

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PODPUNKT "a"
[tex]\large \boxed{3\frac{4}{21} +32\frac{1}{7}=\frac{67}{21}+\frac{225}{7}=\frac{67}{21}+\frac{675}{21}=\frac{742}{21}=35\frac{1}{3}}[/tex]
PODPUNKT "b"
[tex]\large \boxed{5\frac{4}{11}+2\frac{3}{22}- 1\frac{43}{44}=\frac{59}{11}+\frac{47}{22}-\frac{87}{44}=\frac{236}{44}+\frac{94}{44}-\frac{87}{44}=\frac{243}{44}=5\frac{23}{44}}[/tex]
PODPUNKT "c"
[tex]\large \boxed{2,45-1\frac{21}{30}-(-2\frac{3}{25} )=\frac{245}{100}-\frac{51}{30}+\frac{53}{25}=\frac{245}{100}-\frac{170}{100}+\frac{212}{100}=\frac{287}{100}=2\frac{87}{100}}[/tex]PODPUNKT "d"
[tex]\large \boxed{4\frac{1}{3}-2,(3)-3\frac{2}{7}=\frac{13}{3}-\frac{7}{3} -\frac{23}{7} =\frac{6}{3}-\frac{23}{7}=\frac{2}{1}-\frac{23}{7}=\frac{14}{7}-\frac{23}{7}=-\frac{9}{7} =-1\frac{2}{7} }[/tex]