Odpowiedź:
Szczegółowe wyjaśnienie:
a)
[tex]|x-y|=|4\frac{1}{3}-7\frac{3}{4}|=|4\frac{4}{12}-7\frac{9}{12}|=|-3\frac{5}{12}|=3\frac{5}{12}\\\\|x+y||4\frac{1}{3}+7\frac{3}{4}|=|4\frac{4}{12}+7\frac{9}{12}|=|11\frac{13}{12}|=|12\frac{1}{12}|=12\frac{1}{12}[/tex]
b)
[tex]|x-y|=|-5,6-8,2|=|-13,8|=13,8\\\\|x+y|=|-5,6+8,2|=|2,6|=2,6[/tex]
c)
[tex]|x-y|=|-3\frac{3}{7}-(-2\frac{5}{21})|=|-3\frac{3}{7}+2\frac{5}{21}|=|-3\frac{9}{21}+2\frac{5}{21}|=|-1\frac{4}{21}|=1\frac{4}{21}\\\\|x+y|=|-3\frac{3}{7}-2\frac{5}{21}|=|-3\frac{9}{21}-2\frac{5}{21}|=|-5\frac{14}{21}|=5\frac{14}{21}=5\frac{2}{3}[/tex]
d)
[tex]|x-y|=|6\frac{5}{6}-(-10\frac{1}{4})|=|6\frac{5}{6}+10\frac{1}{4}|=|6\frac{10}{12}+10\frac{3}{12}|=|16\frac{13}{12}|=|17\frac{1}{12}|=17\frac{1}{12}\\\\|x+y|=|6\frac{5}{6}-10\frac{1}{4}|=|6\frac{10}{12}-10\frac{3}{12}|=|6\frac{10}{12}-9\frac{15}{12}|=|-3\frac{5}{12}|=3\frac{5}{12}[/tex]
