\(=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\frac{1}{127}\) 

\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\) 

\(=\left(-\frac{18}{18}\right)+\frac{35}{35}+\frac{1}{127}\) 

\(=-1+1+\frac{1}{127}\) 

\(=\frac{1}{127}\)

\(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{127}-\dfrac{7}{18}+\dfrac{4}{35}-\left(-\dfrac{2}{7}\right)\)

\(=\left[-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right]+\left[\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right]+\dfrac{1}{127}\)

\(=-\dfrac{18}{18}+\dfrac{35}{35}+\dfrac{1}{127}\)

\(=-1+1+\dfrac{1}{127}\)

\(=\dfrac{1}{127}\)

\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}-\left(-\frac{2}{7}\right)=-\frac{1}{2}+\frac{3}{5}-\frac{1}{9}+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}+\frac{2}{7}=\frac{1}{127}\)

c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

xcsxccsz

(\(\frac{-1}{2}\)) -(\(\frac{-3}{5}\))+(\(\frac{-1}{9}\))+\(\frac{1}{127}\)-\(\frac{7}{18}\)+\(\frac{4}{35}\)-(\(\frac{-2}{7}\))=\(\frac{448}{1143}\)

a ) 11/125

b ) 1

\(=-\dfrac{1}{127}\)