Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаВопрос:
210. Найдите значение выражения: 17 a) (((-)+6+81-12)-1635)-36; 3.515.5 б) 5 21 28 84 5:1+10 2 15 + 84 2:1+3:1 2 3 1:2+1:3 2 3 57 8:(:(2-1)+:+223 4 14.(5-4)-9+ 3 223; 1 16. 36 35'
Ответ:
Краткое пояснение: Решим примеры, используя порядок действий и правила работы с дробями.
а)
1) \(\,\frac{5}{6}-\frac{3}{8}=\frac{5\cdot4-3\cdot3}{24}=\frac{20-9}{24}=\frac{11}{24}\)
2) \(\,\frac{11}{24}\cdot\frac{8}{33}=\frac{11\cdot8}{24\cdot33}=\frac{1\cdot1}{3\cdot3}=\frac{1}{9}\)
3) \(\,6\frac{35}{84}+8\frac{17}{56}-12\frac{1}{8}=6\frac{5}{12}+8\frac{17}{56}-12\frac{1}{8}=6\frac{70}{168}+8\frac{51}{168}-12\frac{21}{168}=14\frac{121}{168}-12\frac{21}{168}=2\frac{100}{168}=2\frac{25}{42}\)
4) \(\,\frac{1}{9}:2\frac{25}{42}=\frac{1}{9}:\frac{109}{42}=\frac{1}{9}\cdot\frac{42}{109}=\frac{1\cdot14}{3\cdot109}=\frac{14}{327}\)
5) \(\,\frac{11}{1635}-\frac{14}{327}=\frac{11\cdot327-14\cdot1635}{1635\cdot327}=\frac{3597-22890}{534645}=\frac{-19293}{534645}=-\frac{6431}{178215}\)
6) \(\,-\frac{6431}{178215}\cdot36\frac{1}{3}=-\frac{6431}{178215}\cdot\frac{109}{3}=-\frac{6431\cdot109}{178215\cdot3}=-\frac{700979}{534645}=-\frac{700979:3}{534645:3}=-\frac{233659,(6)}{178215}\approx-1,31 \)
1) \(\,\frac{5}{6}-\frac{3}{8}=\frac{5\cdot4-3\cdot3}{24}=\frac{20-9}{24}=\frac{11}{24}\)
2) \(\,\frac{11}{24}\cdot\frac{8}{33}=\frac{11\cdot8}{24\cdot33}=\frac{1\cdot1}{3\cdot3}=\frac{1}{9}\)
3) \(\,6\frac{35}{84}+8\frac{17}{56}-12\frac{1}{8}=6\frac{5}{12}+8\frac{17}{56}-12\frac{1}{8}=6\frac{70}{168}+8\frac{51}{168}-12\frac{21}{168}=14\frac{121}{168}-12\frac{21}{168}=2\frac{100}{168}=2\frac{25}{42}\)
4) \(\,\frac{1}{9}:2\frac{25}{42}=\frac{1}{9}:\frac{109}{42}=\frac{1}{9}\cdot\frac{42}{109}=\frac{1\cdot14}{3\cdot109}=\frac{14}{327}\)
5) \(\,\frac{11}{1635}-\frac{14}{327}=\frac{11\cdot327-14\cdot1635}{1635\cdot327}=\frac{3597-22890}{534645}=\frac{-19293}{534645}=-\frac{6431}{178215}\)
6) \(\,-\frac{6431}{178215}\cdot36\frac{1}{3}=-\frac{6431}{178215}\cdot\frac{109}{3}=-\frac{6431\cdot109}{178215\cdot3}=-\frac{700979}{534645}=-\frac{700979:3}{534645:3}=-\frac{233659,(6)}{178215}\approx-1,31 \)
б)
1) \(\,5:1\frac{1}{2}=5:\frac{3}{2}=5\cdot\frac{2}{3}=\frac{10}{3}=3\frac{1}{3}\)
2) \(\,3\frac{1}{3}+10=13\frac{1}{3}\)
3) \(\,\frac{3}{5}+\frac{5}{21}+\frac{15}{28}+\frac{5}{84}=\frac{3\cdot84+5\cdot20+15\cdot15+5\cdot5}{420}=\frac{252+100+225+25}{420}=\frac{602}{420}=\frac{301}{210}\)
4) \(\,2\frac{1}{2}+3:\frac{1}{3}=\frac{5}{2}+3\cdot\frac{3}{1}=\frac{5}{2}+9=11\frac{1}{2}\)
5) \(\,\frac{301}{210}:13\frac{1}{3}=\frac{301}{210}:\frac{40}{3}=\frac{301}{210}\cdot\frac{3}{40}=\frac{301\cdot1}{70\cdot40}=\frac{301}{2800}\)
6) \(\,\frac{1}{36}-1\frac{16}{35}=\frac{1}{36}-\frac{51}{35}=\frac{35-51\cdot36}{1260}=\frac{35-1836}{1260}=\frac{-1801}{1260}\)
7) \(\,\frac{301}{2800}\cdot\frac{-1801}{1260}=\frac{301\cdot(-1801)}{2800\cdot1260}=\frac{-542101}{3528000}\approx-0,15\)
1) \(\,5:1\frac{1}{2}=5:\frac{3}{2}=5\cdot\frac{2}{3}=\frac{10}{3}=3\frac{1}{3}\)
2) \(\,3\frac{1}{3}+10=13\frac{1}{3}\)
3) \(\,\frac{3}{5}+\frac{5}{21}+\frac{15}{28}+\frac{5}{84}=\frac{3\cdot84+5\cdot20+15\cdot15+5\cdot5}{420}=\frac{252+100+225+25}{420}=\frac{602}{420}=\frac{301}{210}\)
4) \(\,2\frac{1}{2}+3:\frac{1}{3}=\frac{5}{2}+3\cdot\frac{3}{1}=\frac{5}{2}+9=11\frac{1}{2}\)
5) \(\,\frac{301}{210}:13\frac{1}{3}=\frac{301}{210}:\frac{40}{3}=\frac{301}{210}\cdot\frac{3}{40}=\frac{301\cdot1}{70\cdot40}=\frac{301}{2800}\)
6) \(\,\frac{1}{36}-1\frac{16}{35}=\frac{1}{36}-\frac{51}{35}=\frac{35-51\cdot36}{1260}=\frac{35-1836}{1260}=\frac{-1801}{1260}\)
7) \(\,\frac{301}{2800}\cdot\frac{-1801}{1260}=\frac{301\cdot(-1801)}{2800\cdot1260}=\frac{-542101}{3528000}\approx-0,15\)
в)
1) \(\,2\frac{3}{4}-1\frac{15}{28}=\frac{11}{4}-\frac{43}{28}=\frac{11\cdot7-43}{28}=\frac{77-43}{28}=\frac{34}{28}=\frac{17}{14}\)
2) \(\,\frac{17}{14}+\frac{2}{3}=\frac{17\cdot3+2\cdot14}{42}=\frac{51+28}{42}=\frac{79}{42}\)
3) \(\,8:3:\frac{79}{42}=8:\frac{3\cdot79}{42}=8:\frac{237}{42}=8\cdot\frac{42}{237}=\frac{8\cdot14}{79}=\frac{112}{79}\)
4) \(\,\frac{112}{79}+\frac{57}{223}=\frac{112\cdot223+57\cdot79}{79\cdot223}=\frac{24976+4503}{17617}=\frac{29479}{17617}\)
5) \(\,5\frac{5}{7}-4\frac{3}{4}=\frac{40}{7}-\frac{19}{4}=\frac{40\cdot4-19\cdot7}{28}=\frac{160-133}{28}=\frac{27}{28}\)
6) \(\,14\cdot\frac{27}{28}=\frac{14\cdot27}{28}=\frac{1\cdot27}{2}=\frac{27}{2}\)
7) \(\,\frac{27}{2}-9\frac{5}{7}=\frac{27}{2}-\frac{68}{7}=\frac{27\cdot7-68\cdot2}{14}=\frac{189-136}{14}=\frac{53}{14}\)
8) \(\,\frac{53}{14}+\frac{3}{13}=\frac{53\cdot13+3\cdot14}{182}=\frac{689+42}{182}=\frac{731}{182}\)
9) \(\,\frac{29479}{17617}:\frac{731}{182}=\frac{29479}{17617}\cdot\frac{182}{731}=\frac{29479\cdot26}{25167\cdot731}=\frac{766454}{18396477}\approx0,04\)
1) \(\,2\frac{3}{4}-1\frac{15}{28}=\frac{11}{4}-\frac{43}{28}=\frac{11\cdot7-43}{28}=\frac{77-43}{28}=\frac{34}{28}=\frac{17}{14}\)
2) \(\,\frac{17}{14}+\frac{2}{3}=\frac{17\cdot3+2\cdot14}{42}=\frac{51+28}{42}=\frac{79}{42}\)
3) \(\,8:3:\frac{79}{42}=8:\frac{3\cdot79}{42}=8:\frac{237}{42}=8\cdot\frac{42}{237}=\frac{8\cdot14}{79}=\frac{112}{79}\)
4) \(\,\frac{112}{79}+\frac{57}{223}=\frac{112\cdot223+57\cdot79}{79\cdot223}=\frac{24976+4503}{17617}=\frac{29479}{17617}\)
5) \(\,5\frac{5}{7}-4\frac{3}{4}=\frac{40}{7}-\frac{19}{4}=\frac{40\cdot4-19\cdot7}{28}=\frac{160-133}{28}=\frac{27}{28}\)
6) \(\,14\cdot\frac{27}{28}=\frac{14\cdot27}{28}=\frac{1\cdot27}{2}=\frac{27}{2}\)
7) \(\,\frac{27}{2}-9\frac{5}{7}=\frac{27}{2}-\frac{68}{7}=\frac{27\cdot7-68\cdot2}{14}=\frac{189-136}{14}=\frac{53}{14}\)
8) \(\,\frac{53}{14}+\frac{3}{13}=\frac{53\cdot13+3\cdot14}{182}=\frac{689+42}{182}=\frac{731}{182}\)
9) \(\,\frac{29479}{17617}:\frac{731}{182}=\frac{29479}{17617}\cdot\frac{182}{731}=\frac{29479\cdot26}{25167\cdot731}=\frac{766454}{18396477}\approx0,04\)
Ответ: См. решение
