a: Ta có: \(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}-...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

=0

a) \(\dfrac{3}{7}-\dfrac{4}{11}-\dfrac{3}{7}\)

=\(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)-\dfrac{4}{11}\)

\(=-\dfrac{4}{11}\)

b) \(-\dfrac{5}{11}+\dfrac{3}{8}-\dfrac{6}{11}\)

\(=\left(-\dfrac{5}{11}-\dfrac{6}{11}\right)+\dfrac{3}{8}\)

\(=-1+\dfrac{3}{8}=-\dfrac{8}{8}+\dfrac{3}{8}=-\dfrac{5}{8}\)

c) \(\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{11}{4}\)

\(=\left(\dfrac{3}{4}-\dfrac{11}{4}\right)-\dfrac{1}{3}\)

\(=-2-\dfrac{1}{3}=-\dfrac{6}{3}-\dfrac{1}{3}=-\dfrac{7}{3}\)

`a, 3/7 -4/11-3/7`

`=(3/7-3/7)-4/11`

`= 0/7-4/11`

`= -4/11`

__

`b, -5/11 +3/8 -6/11`

`=(-5/11-6/11)+3/8`

`=-11/11 + 3/8`

`= -1+3/8`

`= -8/8+3/8`

`=- 5/8`

__

`c,3/4 -1/3-11/4`

`=(3/4-11/4)-1/3`

`= -8/4-1/3`

`=-2-1/3`

`= -6/3-1/3`

`=-7/3`

có ai làm đc kkkông

a,= 32.(11+67) +78.68

      32.78+78.68

      78.(32+68)

      78.100

      = 7800

b,76.(41-20)+21.24

   =76.21+21.24

   =21.(76+24)

   =21.100

   =2100

a)=-374                   b)=-5250

c)=-2323                  d)=-3069

Có trong SGK ko bro?

a=-97

b=-100

c=-210

d=-200

e=100

a, 64 + -97 + -64

=[64 + (-64)] + (-97)

=0 + (-97)

=-97

a)

\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)                                   

b)

\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)          

d)

\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ =  - 2\end{array}\)

Bài làm 

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(A=\frac{1}{1}-\frac{1}{11}\)

\(A=\frac{10}{11}\)