Question

даю 15 баллов . задание на листочке .

Answer 1

$\displaystyle \tt 2). \ \ \frac{4}{7}=\frac{4\cdot4}{7\cdot4}=\frac{16}{28}\\\\\\{} \ \ \ \ \ \ \frac{5}{7}=\frac{5\cdot4}{7\cdot4}=\frac{20}{28}\\\\\\{} \ \ \ \ \ \ \frac{4}{7}=\frac{16}{28}<\frac{17}{28}<\frac{18}{28}<\frac{19}{28}<\frac{20}{28}=\frac{5}{7};$

$\displaystyle \tt 3).\\a). \ \ \frac{5}{6}+\frac{3}{34}+\frac{2}{51}=\frac{17\cdot5+3\cdot3+2\cdot2}{102}=\frac{85+9+4}{102}=\frac{98}{102}=\frac{49}{51};\\\\\\b). \ \ \frac{3}{10}-\frac{8}{65}-\frac{1}{13}=\frac{3\cdot13-8\cdot2-10}{130}=\frac{39-16-10}{130}=\frac{13}{130}=\frac{1}{10}=0,1;\\\\\\ c). \ \ \bigg(\frac{9}{20}-\frac{3}{8}\bigg)+\frac{11}{24}=\frac{6\cdot9-15\cdot3+11\cdot5}{120}=\frac{54-45+55}{120}=\frac{64}{120}=\frac{8}{15};$

$\displaystyle \tt d). \ \ \bigg(\frac{3}{28}-\frac{2}{35}\bigg)+\bigg(\frac{5}{14}-\frac{4}{21}\bigg)=\frac{5\cdot3-4\cdot2}{140}+\frac{5\cdot3-4\cdot2}{42}=\\\\\\=\frac{7}{140}+\frac{7}{42}=\frac{1}{20}+\frac{1}{6}=\frac{3+10}{60}=\frac{13}{60};$