Вопрос:

6. Ҳисоб кунед: а) \((21-3\frac{7}{16})-(21\frac{5}{12}-\frac{41}{48})\); б) \((3\frac{5}{8}+\frac{1}{4}+2\frac{7}{12})\cdot\frac{9}{56}(4\frac{8}{15}-\frac{11}{3}+\frac{17}{45})\).

Ответ:

а)

\(21-3\frac{7}{16}=\frac{336-55}{16}=\frac{281}{16}\).

\(21\frac{5}{12}-\frac{41}{48}=\frac{1028}{48}-\frac{41}{48}=\frac{987}{48}=\frac{329}{16}\).

Пас, \(\frac{281}{16}-\frac{329}{16}=\frac{-48}{16}=-3\).

б)

\(3\frac{5}{8}+\frac{1}{4}+2\frac{7}{12}=\frac{87}{24}+\frac{6}{24}+\frac{62}{24}=\frac{155}{24}\).

\(4\frac{8}{15}-\frac{11}{3}+\frac{17}{45}=\frac{188}{45}-\frac{165}{45}+\frac{17}{45}=\frac{40}{45}=\frac{8}{9}\).

Пас, \(\frac{155}{24}\cdot\frac{9}{56}\cdot\frac{8}{9}=\frac{155}{168}\).

Ҷавоб: а) \(-3\); б) \(\frac{155}{168}\).