Cześć!
Poziom A
[tex]7\frac{1}{3}+(-6\frac{2}{3})=7\frac{1}{3}-6\frac{2}{3}=6\frac{4}{3}-6\frac{2}{3}=\frac{2}{3}\\\\-5,11+(-3,45)=-(5,11+3,45)=-8,56\\\\-2\frac{5}{6}+(-1\frac{5}{6})=-(2\frac{5}{6}+1\frac{5}{6})=-3\frac{10}{6}=-3\frac{5}{3}=-4\frac{2}{3}\\\\4,2+(-7,8)=4,2-7,8=-(7,8-4,2)=-3,6[/tex]
Poziom B
[tex]8\frac{1}{3}-(-4\frac{2}{3})=8\frac{1}{3}+4\frac{2}{3}=12\frac{3}{3}=13\\\\-5,11-3,45=-(5,11+3,45)=-8,56\\\\2\frac{5}{6}-7\frac{5}{6}=-(7\frac{5}{6}-2\frac{5}{6})=-5\\\\-4,2-(-7,8)=-4,2+7,8=7,8-4,2=3,6[/tex]