\(\dfrac{2}{3}\) . 20.0,375.\(\dfrac{5}{28}\)

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16 tháng 4 2023

7/10 * 8/3 * 20 * 3/8 * 5/28

= (7/10 * 20) * (8/3 * 3/8) * 5/28

= 14 * 1 * 5/28

= 14 * 5/28

= 5/2

16 tháng 4 2023

0,7. 2\(\dfrac{2}{3}\) . 20. 0,375. \(\dfrac{5}{28}\) 

\(\dfrac{14}{15}\) . 20. \(\dfrac{15}{224}\) 

\(\dfrac{56}{3}\) . \(\dfrac{15}{224}\) = \(\dfrac{5}{4}\)

6 tháng 5 2017

\(D=0.7\cdot2\dfrac{2}{3}\cdot20\cdot0.375\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot2\dfrac{2}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\left(\dfrac{8}{3}\cdot\dfrac{3}{8}\right).\left(\dfrac{7}{10}\cdot20\right)\cdot\dfrac{5}{28}\)

\(=\) \(1\) \(\cdot\) \(14\) \(\cdot\) \(\dfrac{5}{28}\)

\(=14\cdot\dfrac{5}{28}\)

\(=\dfrac{5}{2}\).

\(=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{20}{10}\cdot\dfrac{7}{28}\cdot5=2\cdot5\cdot\dfrac{1}{4}=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\)

8 tháng 3 2017

\(A=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{-5}{7}\)

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

\(B=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(=\dfrac{7}{10}.\dfrac{5}{28}.20=\dfrac{5}{2}.\)

8 tháng 3 2017

Ta có :

A = \(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\)

= \(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}+1\)

= \(\left(\dfrac{5}{7}.\dfrac{-2}{11}-\dfrac{5}{7}.\dfrac{9}{11}+\dfrac{5}{7}\right)+1\)

= \(\dfrac{5}{7}.\left(\dfrac{-2}{11}-\dfrac{9}{11}+1\right)+1\)

= \(\dfrac{5}{7}.0+1\)

= 1

B = \(0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

= \(\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

= \(\left(\dfrac{7}{10}.20\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\dfrac{5}{28}\)

= 14.1.\(\dfrac{5}{28}\)

= \(\dfrac{5}{2}\)

Vậy A = 1

B = \(\dfrac{5}{2}\)

17 tháng 4 2017

\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)

\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)

\(A=6-2\dfrac{4}{7}\)

\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)

\(A=3\dfrac{3}{7}\)

\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)

\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)

\(B=2+3\dfrac{7}{11}\)

\(B=5\dfrac{7}{11}\)

\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)

\(C=\dfrac{160}{11}\)

\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)

\(D=\dfrac{7}{28}=\dfrac{5}{2}\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)

\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)

\(\Rightarrow E=0\)

28 tháng 4 2017

Bài 1:

a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)

\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)

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

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

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

b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)

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

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

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

\(=\dfrac{5}{10}+\dfrac{8}{10}\)

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

c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)

\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)

\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)

\(=\dfrac{7}{3}+\dfrac{28}{3}\)

\(=\dfrac{35}{3}\)

d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)

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

\(=\dfrac{1}{6}-\dfrac{21}{6}\)

\(=\dfrac{-10}{3}\)

e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)

\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)

\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)

\(=\dfrac{2}{3}\)

f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)

\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)

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

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

g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)

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

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

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

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

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

h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)

\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)

\(=\dfrac{7}{5}-\dfrac{9}{28}\)

\(=\dfrac{196}{140}-\dfrac{45}{140}\)

\(=\dfrac{151}{140}\)

i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)

\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)

\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)

\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)

k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)

\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)

\(=-\dfrac{2}{3}\)

29 tháng 4 2017

\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)

\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)

\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)

\(A=\dfrac{1}{8}.1.20\)

\(A=\dfrac{20}{8}=\dfrac{5}{2}\)

\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)

\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)

\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)

\(B=\left(16+1\right)+4,03\)

\(B=17+4,03\)

\(B=21,03\)

\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)

\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)

\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)

\(C=390.\dfrac{15}{78}\)

\(C=75\)

22 tháng 3 2017

dễ thế mình làm hết thì tick cho mình nhé

22 tháng 3 2017

uk

Bài 1: 

a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)

c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)

9 tháng 4 2017

a) \(\dfrac{37}{40}-0,64\\ =\dfrac{37}{40}-\dfrac{16}{25}\\ =\dfrac{185}{200}-\dfrac{128}{200}\\ =\dfrac{57}{200}\)

9 tháng 4 2017

b) \(130\dfrac{25}{28}-120\dfrac{12}{35}\\ =\dfrac{3665}{28}-\dfrac{4212}{35}\\ =\dfrac{18325}{140}-\dfrac{16848}{140}\\ =\dfrac{211}{20}\)

2 tháng 4 2017

1. Tìm \(x\):

a) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)

\(\dfrac{x}{5}=\dfrac{1}{5}\)

\(\Rightarrow x=1\)

b) \(\dfrac{-5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}.x\)

\(\dfrac{-5}{6}-\dfrac{7}{12}=x-\dfrac{1}{3}.x\)

\(x-\dfrac{1}{3}.x=\dfrac{-17}{12}\)

\(\dfrac{2}{3}.x=\dfrac{-17}{12}\)

\(x=\dfrac{-17}{12}:\dfrac{2}{3}\)

\(x=\dfrac{-17}{8}\)

c) \(2016^3.2016^x=2016^8\)

\(2016^x=2016^8:2016^3\)

\(2016^x=2016^{8-3}\)

\(2016^x=2016^5\)

\(\Rightarrow x=5\)

d) \(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=3\dfrac{1}{2}\)

\(\left(x+\dfrac{3}{4}\right):\dfrac{5}{2}=\dfrac{7}{2}\)

\(\left(x+\dfrac{3}{4}\right)=\dfrac{7}{2}.\dfrac{5}{2}\)

\(x+\dfrac{3}{4}=\dfrac{35}{4}\)

\(x=\dfrac{35}{4}-\dfrac{3}{4}\)

\(x=\dfrac{32}{4}=8\)

e) \(\left(2,8.x-2^5\right):\dfrac{2}{3}=3^2\)

\(\left(2,8.x-2^5\right)=9.\dfrac{2}{3}\)

\(2,8.x-2^5=6\)

\(2,8.x=6+32\)

\(2,8.x=38\)

\(x=38:2,8\)

\(x=\dfrac{95}{7}\)

f) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{2}{5}\)

\(\dfrac{4}{7}.x=\dfrac{2}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}.x=\dfrac{16}{15}\)

\(x=\dfrac{16}{15}:\dfrac{4}{7}\)

\(x=\dfrac{28}{15}\)

g) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)

\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)

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

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

\(\Rightarrow3x=-6\)

\(x=\left(-6\right):3\)

\(x=-2\)

2 tháng 4 2017

2. Thực hiện phép tính:

a) \(\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}+1\dfrac{4}{5}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{3}+1\right)-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)

\(=\dfrac{1}{2}.\dfrac{5}{3}-\dfrac{1}{3}:\dfrac{3}{4}+\dfrac{9}{5}\)

\(=\dfrac{5}{6}-\dfrac{4}{9}+\dfrac{9}{5}\)

\(=\dfrac{7}{18}+\dfrac{9}{5}\)

\(=\dfrac{197}{90}\)

b) \(\dfrac{7.5^2-7^2}{7.24+21}\)

\(=\dfrac{7.25-7.7}{7.24+7.3}\)

\(=\dfrac{7.\left(25-7\right)}{7.\left(24+3\right)}\)

\(=\dfrac{7.18}{7.27}\)

\(=\dfrac{2}{3}\)

c) \(\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{-4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)

22 tháng 3 2017

\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{13}{4}-\dfrac{7}{-24}\)

\(=\dfrac{85}{24}\)

\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{-3}{56}-\dfrac{3}{28}\)

\(=\dfrac{-9}{56}\)

\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)

\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)

\(=\dfrac{5}{12}\)

\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-4}{7}+\dfrac{4}{7}\)

\(=0\)

\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{47}{24}+\dfrac{5}{-8}\)

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

22 tháng 3 2017

Bài 2 : Tính

a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)

\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)

\(=\dfrac{85}{24}\)

b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)

\(=\dfrac{32+\left(-35\right)-6}{56}\)

\(=\dfrac{-9}{56}\)

c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)

\(=\dfrac{7-32+\left(-24\right)}{36}\)

\(=\dfrac{-49}{36}\)

d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\left(-1\right)+1\)

\(=0\)

e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)

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

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

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