Question

ayudemen porfavor les doy coronita

Answer 1

Respuesta:

Las respuestas respectivamente son: $\frac{3}{8}$ ; $-\frac{1}{6}$ ; $-\frac{13}{10}$ ; $-\frac{164}{35}$

Explicación paso a paso:

a) $- 1 - (-\frac{2}{3} + 1) - [2\frac{1}{4} -(2-\frac{3}{2} )-\frac{2}{6} -\frac{5}{8} ] + 1\frac{3}{2}$

$- 1 - (\frac{-2(1) + 3}{3}) - [\frac{4(2)+1}{4} -(\frac{2(2)-3}{2} )-(\frac{2}{6} +\frac{5}{8}) ] + 1\frac{3}{2}$

$- 1 - (\frac{-2 + 3}{3}) - [\frac{8+1}{4} -(\frac{4-3}{2} )-(\frac{2(8)+6(5)}{6(8)} ) ] + \frac{2(1)+3}{2}$

$- 1 - (\frac{1}{3}) - [\frac{9}{4} -\frac{1}{2} -(\frac{16+30}{48} ) ] + \frac{2+3}{2}$

$- (1 + \frac{1}{3}) - [\frac{9(2)-4(1)}{4(2)}-(\frac{46}{48} ) ] + \frac{5}{2}$

$- (1 + \frac{1}{3}) - [\frac{18-4}{8}-(\frac{23}{24} ) ] + \frac{5}{2}$

$- (\frac{3+1}{3}) - [\frac{14}{8}-\frac{23}{24} ] + \frac{5}{2}$

$- \frac{4}{3} - [\frac{14(24)-8(23)}{8(24)} ] + \frac{5}{2}$

$- \frac{4}{3} - [\frac{336-184}{192} ] + \frac{5}{2}$

$- \frac{4}{3} - [\frac{152}{192} ] + \frac{5}{2}$

$- \frac{4}{3} - [\frac{152}{192} ] + \frac{5}{2}$

$- \frac{4}{3} - \frac{19}{24} + \frac{5}{2}$

$- (\frac{4}{3} + \frac{19}{24}) + \frac{5}{2}$

$- (\frac{4(24)+3(19)}{3(24)}) + \frac{5}{2}$

$- (\frac{96+57}{72}) + \frac{5}{2}$

$- (\frac{153}{72}) + \frac{5}{2}$

$- \frac{17}{8} + \frac{5}{2}$

$\frac{-17(2)+8(5)}{8(2)}$

$\frac{-34+40}{16}$

  $\frac{6}{16}$

$= \frac{3}{8}$

b) $-[3\frac{2}{4} -2,333333.....] + (-\frac{1}{5}) + 1,2$

$-[\frac{4(3)+2}{4} -2,333333.....] -\frac{1}{5}+ \frac{6}{5}$

$-[\frac{12+2}{4} -2,333333.....]+ \frac{6}{5} -\frac{1}{5}$

$-[\frac{14}{4} -2,333333.....]+ \frac{6-1}{5}$

$-[\frac{7}{2} -\frac{23-2}{9}]+ \frac{5}{5}$

$-[\frac{7}{2} -\frac{21}{9}]+ 1$

$-[\frac{7}{2} -\frac{7}{3}]+ 1$

$1-[\frac{7(3)-2(7)}{2(3)}]$

$1-[\frac{21-14}{6}]$

$1-\frac{7}{6}$

$\frac{1(6)-7}{6}$

$\frac{6-7}{6}$

$-\frac{1}{6}$

c) $2\frac{2}{3} -[3,5 - (1 - 0, 6666...)]+1\frac{1}{5} -2$

$\frac{3(2)+2}{3} -[\frac{7}{2} - (1- \frac{6}{9} )]+\frac{5(1)+1}{5} -2$

$\frac{6+2}{3} -[\frac{7}{2} - (\frac{9(1)-6}{9} )]+\frac{5+1}{5} -2$

$\frac{8}{3} -[\frac{7}{2} - (\frac{9-6}{9} )]+\frac{6}{5} -2$

$\frac{8}{3} -[\frac{7}{2} - (\frac{3}{9} )]+\frac{6-5(2)}{5}$

$\frac{8}{3} -[\frac{7}{2} - \frac{1}{3} ]+\frac{6-10}{5}$

$\frac{8}{3} -[\frac{7(3)-2(1)}{2(3)} ]+(\frac{-4}{5})$

$\frac{8}{3} -[\frac{21-2}{6} ]-\frac{4}{5}$

$[\frac{8}{3} -\frac{19}{6}]-\frac{4}{5}$

$[\frac{8(6)-3(19)}{3(6)} ]-\frac{4}{5}$

$[\frac{48-57)}{18} ]-\frac{4}{5}$

$\frac{-9}{18} -\frac{4}{5}$

$-\frac{1}{2} -\frac{4}{5}$

$\frac{-1(5)-2(4)}{2(5)}$

$\frac{-5-8}{10}$

$-\frac{13}{10}$

d. $-1,8-2\frac{1}{4} -[-0,16666...+\frac{1}{3}] -[\frac{4}{3}+(0,333...-0,222...)]+\frac{1}{6}$

$-\frac{9}{5} -\frac{2(4)+1}{4} -[-(0,16666...)+\frac{1}{3}] -[\frac{4}{3}+(0,333...-0,222...)]+\frac{1}{6}$

$-\frac{9}{5} -\frac{8+1}{4} -[-(\frac{16-1}{90} )+\frac{1}{3}] -[\frac{4}{3}+(\frac{3}{9} -\frac{2}{9} )]+\frac{1}{6}$

$-\frac{9}{5} -\frac{9}{4} -[-(\frac{15}{90} )+\frac{1}{3}] -[\frac{4}{3}+(\frac{3-2}{9})]+\frac{1}{6}$

$-(\frac{9}{5} +\frac{9}{4})-[-\frac{1}{6}+\frac{1}{3}] -[\frac{4}{3}+\frac{1}{9}]+\frac{1}{6}$

$-(\frac{9(4)+5(9)}{5(4)})-[\frac{-1(3)+6(1)}{6(3)}] -[\frac{4(9)+3(1)}{3(9)}]+\frac{1}{6}$

$-(\frac{36+45}{20})-[\frac{-3+6}{18}] -[\frac{36+3}{27}]+\frac{1}{6}$

$-\frac{81}{20}-\frac{3}{18} -\frac{39}{27}+\frac{1}{6}$

$-(\frac{81}{20}+\frac{1}{6} )-\frac{13}{9}+\frac{1}{6}$

$-(\frac{81(6)+20(1)}{20(6)} )-\frac{13}{9}+\frac{1}{6}$

$-(\frac{486+20}{120} )-\frac{13}{9}+\frac{1}{6}$

$-(\frac{506}{120} )-\frac{13}{9}+\frac{1}{6}$

$-\frac{253}{60} -\frac{13}{9}+\frac{1}{6}$

$-(\frac{253}{60} +\frac{13}{9})+\frac{1}{6}$

$-(\frac{253(9)+60(13)}{60(9)})+\frac{1}{6}$

$-(\frac{2277+780}{630})+\frac{1}{6}$

$-(\frac{3057}{630})+\frac{1}{6}$

$-\frac{1019}{210}+\frac{1}{6}$

$\frac{-1019(6)+210(1)}{210(6)}$

$\frac{-6114+210}{1260}$

$\frac{-5904}{1260}$

$-\frac{164}{35}$

Psdt: Agradecería demasiado si pudieras puntuar mi respuesta y marcarla con la coronita.

Si está solución puede ayudar a más personas, un gracias me ayudaría mucho. =)