Question

ayuda plis con su procedimiento doy coronita ​

Answer 1

Explicación paso a paso:

a)

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

$-\frac{5}{8} +\frac{1}{1} =\frac{-5+8}{8} =\frac{3}{8}$

b)

$-\frac{2}{7} +\frac{-5}{14}$

$-\frac{2}{7} +\frac{-5}{14}=\frac{-4+(-5)}{14}=\frac{-9}{14}$

c)

$-\frac{2}{9} +\frac{5}{9} +3$

$-\frac{2}{9} +\frac{5}{9} +\frac{3}{1} =\frac{-2+5+27}{9} =\frac{30}{9} =\frac{10}{3}$

d)

Suponiendo que el segundo término es una fracción mixta:

$\frac{3}{2}+2\frac{1}{3}+\frac{4}{5}+\frac{0}{6}$

$\frac{3}{2}+\frac{7}{3}+\frac{4}{5}+0=\frac{15*3+10*7+6*4}{30} =\frac{139}{30}$

e)

$-\frac{3}{2} -\frac{2}{5}$

$-\frac{3}{2} -\frac{2}{5} =\frac{-5*3-2*2}{10}=-\frac{19}{10}$

f)

$-\frac{4}{7}-\frac{5}{7}-\frac{7}{7}$

$-\frac{4}{7}-\frac{5}{7}-\frac{1}{1}=\frac{-4-5-7}{7}=-\frac{16}{7}$

g)

$\frac{1}{6}-(\frac{1}{5}-2)$

$\frac{1}{6}-\frac{1}{5}+\frac{2}{1}=\frac{5-6+2*30}{30} =\frac{59}{30}$

h)

$\frac{2}{3}+\frac{1}{5}-\frac{4}{15}$

$\frac{2}{3}+\frac{1}{5}-\frac{4}{15} =\frac{5*2+3*1-4}{15}=\frac{9}{15} =\frac{3}{5}$

Multiplicación y División

a)

$4*\frac{6}{7} = \frac{4}{1} *\frac{6}{7}=\frac{4*6}{7}=\frac{24}{7}$

b)

$\frac{2}{5}*\frac{5}{2} =\frac{10}{10}=1$

c)

$-\frac{11}{6}*\frac{10}{4}=\frac{-11*10}{6*4}=\frac{-110}{24}=-\frac{55}{12}$

d)

$\frac{1}{8}*(-\frac{7}{4} )=\frac{1*(-7)}{8*4}=-\frac{7}{32}$

e)

$\frac{2}{5} : \frac{6}{10}*\frac{3}{8}$

$\frac{2*10}{5*6} *\frac{3}{8}$

$\frac{20}{30} *\frac{3}{8}$

$\frac{2}{3} *\frac{3}{8}=\frac{2*3}{3*8} =\frac{6}{24} =\frac{1}{4}$

f)

$3\frac{1}{3}*(-\frac{4}{5}):\frac{16}{15}$

$\frac{10}{3}*(-\frac{4}{5}):\frac{16}{15}$

$\frac{10*(-4)}{3*5}:\frac{16}{15}$

$\frac{-40}{15}:\frac{16}{15}$

$\frac{-8}{3}:\frac{16}{15}=\frac{(-8)*15}{16*3} =\frac{-120}{48}=-\frac{5}{2}$

g)

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

$\frac{1}{2} -\frac{1}{4}=\frac{2*1-1}{4}=\frac{1}{4}$

$1\frac{1}{2} -\frac{1}{5}=\frac{3}{2}-\frac{1}{5}=\frac{5*3-2*1}{10}=\frac{13}{10}$

$\frac{1}{4} *\frac{13}{10} =\frac{1*13}{4*10} =\frac{13}{40}$

h)

$4+\frac{2}{7} :\frac{6}{7}$

$4+(\frac{2*7}{7*6})=4+\frac{14}{42} =\frac{4}{1} +\frac{1}{3}=\frac{3*4+1}{3}=\frac{13}{3}$

i)

$(2+\frac{1}{5})*(1+\frac{2}{3})$

$2+\frac{1}{5}=\frac{2}{1} +\frac{1}{5}=\frac{5*2+1}{5}=\frac{11}{5}$

$1+\frac{2}{3}=\frac{1}{1}+\frac{2}{3} =\frac{3+2}{3}=\frac{5}{3}$

$\frac{11}{5} *\frac{5}{3} =\frac{11*5}{5*3}=\frac{55}{15}=\frac{11}{3}$

j)

$(\frac{3}{4} -\frac{1}{3} ):( 1+\frac{1}{2})$

$\frac{3}{4} -\frac{1}{3}=\frac{3*3-4*1}{12}=\frac{5}{12}$

$1+\frac{1}{2}= \frac{1}{1} +\frac{1}{2}=\frac{2*1+1*1}{2}=\frac{3}{2}$

$\frac{5}{12}:\frac{3}{2} =\frac{5*2}{12*3}=\frac{10}{36}=\frac{5}{18}$

k)

$\frac{9}{11}*(2\frac{3}{7}-1\frac{5}{9} )$

$\frac{9}{11}*(\frac{17}{7}-\frac{14}{9} )$

$\frac{17}{7}-\frac{14}{9} =\frac{9*17-7*14}{63}=\frac{55}{63}$

$\frac{9}{11}*\frac{55}{63}=\frac{9*55}{11*63}=\frac{495}{693}=\frac{5}{7}$

l)

$\frac{13}{6}*\frac{22}{39} +\frac{5}{13}$

$\frac{13}{6}*\frac{22}{39} =\frac{13*22}{6*39} =\frac{286}{234}=\frac{11}{9}$

$\frac{11}{9} +\frac{5}{13}=\frac{11*13+5*9}{117}=\frac{188}{117}$