A=\(\left[\dfrac{\dfrac{42}{31}.\dfrac{31}{7}-\left(15-\dfrac{2}{3}\right)}{\dfrac{29}{6}+\dfrac{1}{6}.\dfrac{20}{3}}\right].\dfrac{31}{50}\)

= \(\left(\dfrac{6-\dfrac{43}{3}}{\dfrac{29}{6}+\dfrac{10}{9}}\right).\dfrac{31}{50}\)=\(\left(\dfrac{\dfrac{-25}{3}}{\dfrac{107}{18}}\right).\dfrac{31}{50}\)=\(\dfrac{-150}{107}.\dfrac{31}{50}\)=\(\dfrac{-93}{107}\)

\(=\dfrac{\dfrac{142}{31}\cdot\dfrac{31}{7}-\left(\dfrac{3}{2}-\dfrac{19}{3}\cdot\dfrac{2}{19}\right)}{\dfrac{29}{6}+\dfrac{1}{6}\cdot\dfrac{20}{3}}\cdot\dfrac{-93}{14}\)

\(=\dfrac{\dfrac{142}{7}-\dfrac{3}{2}+\dfrac{2}{3}}{\dfrac{29}{6}+\dfrac{10}{9}}\cdot\dfrac{-93}{14}\)

\(=\dfrac{817}{42}:\dfrac{107}{18}\cdot\dfrac{-93}{14}\cong-21,74\)

a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)

\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)

b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)

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

c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)

\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)

\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)

\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)=\dfrac{7}{8}.\left(\dfrac{10}{60}+\dfrac{24}{60}\right)=\dfrac{7}{8}.\dfrac{17}{30}=\dfrac{114}{240}\)

b)\(\dfrac{3}{2}-\dfrac{5}{6}\left(\dfrac{1}{2}\right)^2+\sqrt{4}=\dfrac{3}{2}-\dfrac{5}{6}.\dfrac{1}{4}+2=\dfrac{3}{2}-\dfrac{5}{24}+2=\dfrac{36}{24}-\dfrac{5}{24}+\dfrac{48}{24}=\dfrac{79}{24}\)c)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-1\dfrac{15}{17}+\dfrac{2}{3}=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}=1+\left(\dfrac{7}{21}+\dfrac{14}{21}\right)-\dfrac{32}{17}=1+1-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{34}{17}-\dfrac{32}{17}=\dfrac{2}{17}\)d)\(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)=-8.\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-8.\dfrac{2}{4}:\left(\dfrac{54}{24}-\dfrac{28}{24}\right)=-8.\dfrac{2}{4}:\dfrac{13}{12}=-4.\dfrac{12}{13}=\dfrac{-48}{13}\)e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)+28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)=\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right).\left(-\dfrac{5}{3}\right)=\left(\dfrac{120}{7}+\dfrac{196}{7}\right).\left(-\dfrac{5}{3}\right)=\dfrac{316}{7}.\left(-\dfrac{5}{3}\right)=-\dfrac{1580}{21}\)

bài này tự almf

a) 119/240

b)1/6

c)2/17

d)-48/13

e)-520/7

a: \(=\dfrac{7}{8}\cdot\dfrac{10+24}{60}=\dfrac{7}{8}\cdot\dfrac{34}{60}=\dfrac{7}{8}\cdot\dfrac{17}{30}=\dfrac{119}{240}\)

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

c: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)-1-\dfrac{15}{17}=1-\dfrac{15}{17}=\dfrac{2}{17}\)

d: \(=\left(-8\right)\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

=-4:(27/12-14/12)

=-4:13/12=-4*12/13=-48/13

e: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)=-\dfrac{520}{7}\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)

2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)

c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)

\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)

\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

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

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)