a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)

\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)

b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)

\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)

c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)

\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)

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

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

\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)

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

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

b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)

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

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

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

\(=2\dfrac{3}{7}\)

c,

= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)

= \(\dfrac{5}{9}.1\)

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

a,

= \(\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{-4}{3}\)

= \(\dfrac{1}{3}.\dfrac{10}{5}+\dfrac{-4}{3}\)

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

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

a) \(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(-\dfrac{5}{6}x=\dfrac{5}{12}\)

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

b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3.7\right)=-\dfrac{53}{10}\)

\(\dfrac{3}{5}\left(3x-3.7\right)=-\dfrac{57}{10}\)

\(3x-3.7=-\dfrac{19}{2}\)

\(3x=-5.8\)

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

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}x\right)=\dfrac{8}{27}\)

\(2+\dfrac{3}{4}x=\dfrac{21}{8}\)

\(\dfrac{3}{4}x=\dfrac{5}{8}\)

\(x=\dfrac{5}{6}\)

d) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

\(x=-\dfrac{3}{20}\)

\(\dfrac{2}{3}x-\dfrac{3}{2}x=\dfrac{5}{12}\)

\(\left(\dfrac{2}{3}-\dfrac{3}{2}\right)x=\dfrac{5}{12}\)

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

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

a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)

b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)

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

c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)

\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)

d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)

\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)

\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)

\(D=1+-1+\dfrac{1}{41}\)

\(D=0+\dfrac{1}{41}\)

\(D=\dfrac{1}{41}\)

\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)

=1/57

\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)

ta có

\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)

_______________________ X ________________________

\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)

= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)

a) \(\dfrac{8}{5}-\dfrac{9}{5}=\dfrac{8-9}{5}=\dfrac{-1}{5}\)

b) \(\dfrac{5}{2}+\dfrac{2}{3}=\dfrac{15}{6}+\dfrac{4}{6}=\dfrac{15+4}{6}=\dfrac{19}{6}\)

c) \(\dfrac{-5}{9}\cdot\dfrac{2}{11}=\dfrac{-5\cdot2}{9\cdot11}=\dfrac{-10}{99}\)

d) \(\dfrac{-2}{9}:\dfrac{1}{3}=\dfrac{-2}{9}\cdot3=\dfrac{-2}{3}\)

e) \(\dfrac{3}{8}-\dfrac{1}{4}+\dfrac{5}{12}=\dfrac{9}{24}-\dfrac{6}{24}+\dfrac{10}{24}=\dfrac{9-6+10}{24}=\dfrac{13}{24}\)

f) \(\dfrac{-4}{3}\cdot\dfrac{5}{4}:\dfrac{7}{3}=\dfrac{-4}{3}\cdot\dfrac{5}{4}\cdot\dfrac{3}{7}=\dfrac{-4\cdot5\cdot3}{3\cdot4\cdot7}=\dfrac{-5}{7}\)

a) \(\left(6\dfrac{1}{9}+3\dfrac{7}{11}\right)-1\dfrac{1}{9}\)

\(=\left(\left(6+3\right)+\left(\dfrac{1}{9}+\dfrac{7}{11}\right)\right)-\dfrac{10}{9}\)

\(=\left(9+\dfrac{74}{99}\right)-\dfrac{10}{9}\)

\(=9\dfrac{74}{99}-\dfrac{10}{9}\)

\(=\dfrac{965}{99}-\dfrac{10}{9}\)

\(=\dfrac{95}{11}\)

b) \(1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:5\%\)

\(=1\dfrac{1}{3}-2\dfrac{5}{6}-\dfrac{6}{11}+3:\dfrac{1}{20}\)

\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+3\cdot20\)

\(=\dfrac{4}{3}-\dfrac{17}{6}-\dfrac{6}{11}+60\)

\(=\dfrac{1275}{22}\)

c) \(4\dfrac{3}{4}-0,37+\dfrac{1}{8}-1,28-2,5+3\cdot\dfrac{1}{2}\)

\(=\dfrac{19}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{3}{2}\)

\(=\dfrac{89}{40}\)

d) \(\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{11}+\dfrac{10}{77}-5\cdot\dfrac{2}{11}\)

\(=\dfrac{5}{11}+\dfrac{10}{77}-\dfrac{10}{11}\)

\(=-\dfrac{25}{77}\)

e) \(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

(câu này chờ mình một chút)

\(a)\left(2\dfrac{5}{6}+1\dfrac{4}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{6}+\dfrac{13}{9}\right):\left(10\dfrac{1}{12}-9\dfrac{6}{12}\right)\)

\(=\left(\dfrac{153}{54}+\dfrac{78}{54}\right):\left(1\dfrac{-5}{12}\right)\)

\(=\dfrac{231}{54}:\dfrac{7}{12}\)

\(=\dfrac{198}{27}\)

\(b)\dfrac{0,8\left(\dfrac{4}{5}:1,25\right)}{0,64-\dfrac{1}{25}}\)

\(=\dfrac{0,8\left(0,8:1,25\right)}{0,64-0,04}\)

\(=\dfrac{0,8.0,64}{0,6}\)

\(=\dfrac{0,512}{0,6}\)\(=\dfrac{64}{75}\)