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)

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

a: =17-70-163+246

=-43-163+246

=246-206

=40

có ai làm đc kkkông

a) \(\dfrac{-5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{2}{9}\right)-\left(\dfrac{4}{15}+\dfrac{11}{15}\right)\)

\(=\dfrac{-3}{9}-\dfrac{15}{15}\)

\(=-\dfrac{1}{3}-1\)

\(=-\dfrac{4}{3}\)

b) \(\dfrac{-8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{6}\)

\(=\left(\dfrac{-8}{13}+\dfrac{21}{13}\right)-\dfrac{7}{16}-\dfrac{1}{6}\)

\(=\dfrac{13}{13}-\dfrac{29}{48}\)

\(=1-\dfrac{29}{48}\)

\(=\dfrac{19}{48}\)

c) \(\dfrac{-15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\dfrac{-3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\left(\dfrac{-3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)

\(=\dfrac{5}{5}-\dfrac{7}{3}\)

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

\(=-\dfrac{4}{3}\)

a: =-5/9+2/9-11/15-4/15

=-3/9-1

=-12/9=-4/3

b: =-8/13+21/13-7/16-1/6

=1-1/6-7/16

=5/6-7/16

=19/48

c: =-15/25+8/5-2/3-5/3

=-7/3+1

=-4/3

a) \(-\dfrac{5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)

\(=\left(-\dfrac{5}{9}+\dfrac{2}{9}\right)+\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)\)

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

b) \(-\dfrac{8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{16}\)(sửa đề)

\(=\left(-\dfrac{8}{13}+\dfrac{21}{13}\right)+\left(-\dfrac{7}{16}-\dfrac{1}{16}\right)\)

\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)

c) \(-\dfrac{15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=-\dfrac{3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\left(-\dfrac{3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)

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

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`

\(a,12.\dfrac{-7}{11}.\dfrac{5}{6}.\dfrac{22}{7}=\left(12.\dfrac{5}{6}\right)\left(\dfrac{-7}{11}.\dfrac{22}{7}\right)=10.\left(-2\right)=-20\\ b,\dfrac{-8}{15}.\dfrac{7}{9}.\dfrac{5}{8}.\left(-18\right)=\left(\dfrac{-8}{15}.\dfrac{5}{8}\right)\left[\dfrac{7}{9}.\left(-18\right)\right]=\dfrac{-1}{3}.\left(-14\right)=\dfrac{14}{3}\)

a) 12.−711.56.227= (12.56)(−711.227)= 10.(−2)= −20.

b) −815.79.58.(−18)= (−815.58)[79.(−18)]= −13.(−14)= 143.

a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)

\(=1-\dfrac{29}{15}\)

\(=\dfrac{15-29}{15}\)

\(=-\dfrac{14}{15}\) 

b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)

\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=\dfrac{7}{9}\cdot1\)

\(=\dfrac{7}{9}\)

c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\) 

\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)

\(=\dfrac{25}{63}\)

d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)

\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)

\(=3\cdot\dfrac{9}{11}\)

\(=\dfrac{27}{11}\)

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)