\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right)\cdot\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}+\frac{3}{35}\right)\cdot\left(\frac{-4}{3}\right)}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right)\cdot\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right)\cdot\left(\frac{-4}{3}\right)}=\frac{\left(\frac{-19}{60}\right)\cdot\frac{5}{19}}{\frac{3}{10}\cdot\left(\frac{-4}{3}\right)}\)

\(=\frac{\frac{-1}{12}}{\frac{-2}{5}}=\left(\frac{-1}{12}\right):\left(\frac{-2}{5}\right)=\frac{5}{24}\)

cô choa cụng cho bài ni

Ta có \(\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}=\left(\frac{3}{10}-\frac{6}{20}-\frac{4}{15}-\frac{1}{20}\right).\frac{5}{19}=-\left(\frac{4}{15}+\frac{1}{20}\right).\frac{5}{19}\)

\(=-\frac{1}{5}\left(\frac{4}{3}+\frac{1}{4}\right).\frac{5}{19}=-\left(\frac{4}{3}+\frac{1}{4}\right).\frac{1}{19}=-\frac{19}{12}.\frac{1}{19}=-\frac{1}{12}\)

Lại có \(\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right).\frac{-4}{3}=\left(\frac{1}{14}+\frac{2}{35}\right).\frac{-4}{3}\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{21}=\frac{7}{10}.\frac{-4}{21}=\frac{-2}{15}\)

Khi đó \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{3}{35}\right).\frac{-4}{3}}=\frac{\frac{1}{12}}{-\frac{2}{15}}=-\frac{5}{8}\)

Làm lại : 

\(\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{1}{7}-\frac{5}{35}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)=\left(\frac{1}{14}+\frac{2}{35}\right)\left(-\frac{4}{3}\right)\)

\(=\frac{1}{7}\left(\frac{1}{2}+\frac{2}{5}\right).\frac{-4}{3}=\frac{-4}{21}.\frac{9}{10}=\frac{-6}{35}\)

Khi đó A = \(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{-\frac{1}{12}}{-\frac{6}{35}}=\frac{35}{72}\)

ta có :

\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}=\frac{\frac{1}{5}\left(\frac{3}{2}-\frac{2}{3}-\frac{7}{4}\right).\frac{5}{19}}{\frac{1}{7}\left(\frac{1}{2}+1+\frac{3}{5}\right).\frac{-4}{3}}=\frac{\frac{1}{19}\left(-\frac{2}{3}-\frac{1}{4}\right)}{-\frac{3}{21}\left(\frac{3}{2}+\frac{3}{5}\right)}=\frac{\frac{1}{19}.\frac{-11}{12}}{-\frac{3}{21}.\frac{21}{10}}\)

\(=\frac{\frac{1}{19}.\frac{-11}{12}}{-\frac{3}{10}}=\frac{11}{19.12}.\frac{10}{3}=\frac{55}{342}\)

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

a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)

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

= (-1) + \(\dfrac{3}{4}\)

= \(\dfrac{-4}{4}+\dfrac{3}{4}\)

= \(\dfrac{-1}{4}\)

b; 0,5  + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))

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

= 1 + 1 

= 2

c; \(\dfrac{1}{3}\) + \(\dfrac{-1}{5}\) - \(\dfrac{-1}{4}\) - \(\dfrac{7}{9}\)- \(\dfrac{-5}{12}\) + \(\dfrac{-1}{45}\) + \(\dfrac{1}{41}\)

  = (\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{1}{5}\) + \(\dfrac{7}{9}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{9}{45}+\dfrac{35}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= (\(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)) - (\(\dfrac{44}{45}\) + \(\dfrac{1}{45}\)) + \(\dfrac{1}{41}\)

= 1 - 1 + \(\dfrac{1}{41}\)

= 0 + \(\dfrac{1}{41}\)

= \(\dfrac{1}{41}\)