a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)

b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)

c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)

d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)

\(\frac{-11}{23}\cdot\frac{6}{7}+\frac{8}{7}\cdot\frac{-11}{23}-\frac{1}{23}\)

\(=\frac{-11}{23}\cdot\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}\cdot\frac{14}{7}-\frac{1}{23}\)

\(=\frac{-11}{23}\cdot2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=\frac{-22-1}{23}\)

\(=\frac{-23}{23}=-1\)

đáp án =(-1)

=-11/23(6/7+8/7)-1/23

=-11/23.2-1/23

=-22/23-1/23

=-1

****

1)

\(2.\frac{3}{7}+\left(\frac{2}{9}-1\frac{3}{7}\right)-\frac{5}{3}:\frac{1}{9}\)

\(=\frac{6}{7}+\frac{2}{9}-\frac{10}{7}-\frac{5}{3}.9\)

\(=\left(\frac{6}{7}-\frac{10}{7}\right)+\frac{2}{9}-15\)

\(=\frac{-4}{7}+\frac{2}{9}-\frac{15}{1}\)

\(=\frac{-36}{63}+\frac{14}{63}-\frac{945}{63}\)

\(=\frac{-967}{63}\)

2)

\(-\frac{11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)

\(=\frac{-11}{23}\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=\frac{-23}{23}\)

\(=-1\)

3)

\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{4}{24}-\frac{3}{24}-\frac{1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\frac{0}{24}\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).0\)

\(=0\)

a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)

\(=\frac{15-4}{26}=\frac{11}{26}\)

c) \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)

d) Đặt \(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}=A\)

Ta có: \(A.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)=A.\left(\frac{4-3-1}{24}\right)=A.0=0\)

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)

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

=-11/23x(6/7+8/7)-1/23

=-11/23x2-1/23

=-22/23-1/23

=-1

\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}\)=\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)=\(\frac{-11}{23}.2-\frac{1}{23}\)=\(\frac{-22}{23}-\frac{1}{23}=\frac{-21}{23}\)

bạn Nguyễn Thu Dương sai ở chỗ -22/23-1/23=-22/23+-1/23 chứ

a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\)

C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\)

a.

\(-\frac{3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)\)

\(=-\frac{3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{3}{7}-\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)

\(=0+\left(\frac{15-4}{26}\right)\)

\(=\frac{11}{26}\)

c.

\(-\frac{11}{23}\times\frac{6}{7}+\frac{8}{7}\times\frac{-11}{23}-\frac{1}{23}\)

\(=-\frac{11}{23}\times\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=-\frac{11}{23}\times\frac{14}{7}\)

\(=-\frac{11}{23}\times2-\frac{1}{23}\)

\(=-\frac{22}{23}-\frac{1}{23}\)

\(=-\frac{23}{23}\)

\(=-1\)

d.

\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right)\times\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right)\times\left(\frac{4-3-1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right)\times\frac{0}{24}\)

\(=0\)

Chúc bạn học tốt

Ta có: \(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}=\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}=\frac{7}{9}\)\(\frac{7}{9}\)

Chúc e học tốt!