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1)
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{6}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)
\(=\dfrac{3}{-5}+\dfrac{3}{5}\)
\(=-\dfrac{3}{5}+\dfrac{3}{5}\)
\(=0\)
\(a)0,2-1\dfrac{3}{7}-\dfrac{6}{5}\)
\(=\dfrac{1}{5}-\dfrac{10}{7}-\dfrac{6}{5}\)
\(=(\dfrac{1}{5}-\dfrac{6}{5}\)-\dfrac{10}{7}\)
\(=-1-\dfrac{10}{7}\)
\(=\dfrac{-17}{7}\)
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\(b)(\dfrac{4}{5}-1):\dfrac{3}{5}-\dfrac{2}{3}.0,5\)
\(=\dfrac{-1}{5}.\dfrac{5}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{-1}{3}-\dfrac{1}{3}=\dfrac{-2}{3}\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}+\dfrac{6}{7}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}+0,875+0,7}+\dfrac{6}{7}.\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}.\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1.\)
Vậy.....
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
a, \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\)
\(\Rightarrow\dfrac{x}{12}=\dfrac{11}{12}\)
\(\Rightarrow x=11\)
b, \(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{15}\)
\(\Rightarrow x=1\)
c, \(-2^3+0,5x=1,5\)
\(\Rightarrow-8+\dfrac{1}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{2}+8\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{19}{2}\)
\(\Rightarrow x=19\)
1) \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\) 2)\(\dfrac{2}{3}-1\dfrac{4}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}\) \(\dfrac{2}{3}-\dfrac{19}{15}x=\dfrac{-3}{5}\) \(\dfrac{x}{12}=\dfrac{11}{12}\) \(\dfrac{19}{15}x=\dfrac{2}{3}-\left(\dfrac{-3}{5}\right)\) => \(x=11\) \(\dfrac{19}{15}x=\dfrac{19}{15}\) => \(x=1\) 3) -23 + 0,5x = 1,5 -8 + 0,5x = 1,5 0,5x = 1,5 - (-8) 0,5x = 9,5 x = 9,5 : 0,5 x = 19
1: \(=\dfrac{-8}{11}\left(\dfrac{3}{2}+\dfrac{33}{20}+\dfrac{11}{10}\right)\)
\(=\dfrac{-8}{11}\cdot\dfrac{30+33+22}{20}=\dfrac{-8}{11}\cdot\dfrac{85}{20}=-\dfrac{34}{11}\)
2: \(=\dfrac{2}{3}+\dfrac{1}{3}=1\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)
\(M=0\times\dfrac{2022}{2021}\)
M=0
a: \(\Leftrightarrow-x+2=-0.02\)
=>2-x=-0,02
=>x=2,02
b: \(\Leftrightarrow-0.75x+9=0.4\)
=>-0,75x=-8,6
=>x=172/15
c: \(\Leftrightarrow0.27x-5.4=-0.375\)
=>0,27x=5,025
=>x=335/18